Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C.

Other Exercises

Question 1.
Show that:
(i) tan 10° tan 15° tan 75° tan 80° = 1
(ii) sin 42° sec 48°+cos 42° cosec 48°= 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q1.1
Solution:
(i) tan 10° tan 15° tan 75° tan 80°= 1
L.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan (90° – 80°) tan (90° – 75°) tan 75° tan 80°
= cot 80° cot 75° tan 75° tan 80°
= tan 80° cot 80° x tan 75° cot 75°
= 1 x 1 = 1= R.H.S. (∵ tan A cot A = 1)
(ii) sin 42° sec 48°+ cos 42° cosec 48°= 2
L.H.S. = sin 42° sec 48°+ cos 42° cosec 48°
= sin 42° sec (90° – 42°) + cos 42° cosec (90° – 42°)
= sin 42° cosec 42°+ cos 42° sec 42°
=1 + 1=2 R.H.S. (∵ sin A cosec A=1, cos A sec A=1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q1.2

Question 2.
Express each of the following in terms of angles between 0°and 45°.
(i) sin 59° + tan 63°
(ii) cosec 68° + cot 72°
(iii) cos 74° + sec 67°
Solution:
(i) sin 59° + tan 63°
= sin (90° – 31°) + tan (90° – 27°)
= cos 31°+ cot 27°
(ii) cosec 68° + cot 72°
= cosec (90° – 22°) + cot (90° – 18°)
= sec 22°+ tan 18°
(iii) cos 74°+ sec 67°
= cos (90° – 16°) + sec (90° – 23°)
= sin 16°+ cosec 23°

Question 3.
Show that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q3.3

Question 4.
For triangle ABC, Show that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q4.2

Question 5.
Evaluate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.5

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q6.2

Question 7.
Find (in each case, given below) the value of x, if:
(i) sin x = sin 60° cos 30° – cos 60° sin 30°
(ii) sin x = sin 60° cos 30° + cos 60° sin 30°
(iii) cos x = cos 60° cos 30° – sin 60° sin 30°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.1
(v) sin 2x = 2 sin 45° cos 45° 
(vi) sin 3x = 2 sin 30° cos 30°
(vii) cos (2x – 6°) = cos2 30° – cos2 60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.5

Question 8.
In each case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
(i) sin (90° – 3A). cosec 42° = 1
(ii) cos (90° – A). sec 77° = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q8.1

Question 9.
Prove that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q9.2

Question 10.
Evaluate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q10.2

Question 11.
Without using trigonometric tables, evaluate sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q11.1

Question 12.
Without using trigonometrical tables, evaluate: cosec2 57° – tan2 33° + cos 44° cosec 46° – \( \sqrt{2} \) cos45°- tan2 60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q12.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B.

Other Exercises

Question 1.
Prove that: 
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.2

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.4

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.6

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.7

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.9

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.10
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.11

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.12

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.13
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.14

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.15
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.17

Question 2.
If xcosA + ysinA = m and xsinA-ycosA = n, then prove that: x2+y2 = m2 + n2
Solution:
x cos A + y sin A = m    …(i)
x sin A – y cos A = n     ….(ii)
squaring (i) and (ii)
x2 cos2 A + y2 sin2 A + 2 xy cosA sinA = m2
x2 sin2 A + y2 cos2 A – 2 xy cos A sin A = n2
Adding we get,
x2 (sin2 A + cos2 A) + y2 (sin2 A + cos2 A) = m2+n2
∴ x2+y2 = m2 + n2(∵ sin2A + cos2A= 1)
Hence proved.

Question 3.
If m = a sec A +b tan A and n=atanA + bsecA, then prove that: m2-n2 = a2-b2
Solution:
m = asec A + btan A         ……(i)
n = a tan A + b sec A       …..(ii)
squaring (i) and (ii)
m2 = a2 sec2 A + b2 tan2 A + 2ab sec A tan A
n2 = a2 tan2 A + b2 sec2 A + 2 ab tan A sec A
Subtracting, we get
m2 – n2 = a2 (sec2 A – tan2 A) + b2 (tan2 A – sec2 A)
= a2x 1 +b2(-1) = a2-b2 ( ∵ sec2A-tan2A= 1)  .
Henceproved

Question 4.
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that: x2 + y2 + z2 = r.
Solution:
x = r sin A cos B      ….(i)
y = r sin A sin B      ….(ii)
z = r cosA               …….(iii)
Squaring, (i), (ii) & (iii)
x2=r2 sin2 A cos2 B,
y2 = r2sin2Asin2B,
z2 = r2cos2A
Adding, we get,
x2+y2 + z2=r2 (sin2A cos2E + sin2 A sin2 B+cos2A)
= r[sin2A (cos2 B + sin2B) + cos2A]
= r [sin2 A x 1 + cos2 A]
= r2 [sin2 A + cos2 A] = r2 x 1  = r2        ( ∵ sin2 A + cos2 A = 1)
Hence proved.

Question 5.
If sin A + cos A = m and sec A + cosec A=n, show that n (m2-1) = 2m
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q5.1

Question 6.
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that x2 + y2 + z2 = r2
Solution:
x = r cosAcosB              ….(i)
y = r cosAsinB             ….(ii)
z = r sinA                 ….(iii)
Squaring (i), (ii), (iii)
x2 = r2 cos2 A cos2 B, y2 = r2 cos2 A sin2B
z2 = r2sin2A
Adding, we get
x2 + y2 + z2 = r2 (cos2 A cos2B + cos2 A sin2 B + sin2 A)
= r2 [cos2 A (cos2 B + sin2B) + sin2 A]
= r2[cos2Ax 1+sin2A]
= r2 (1) = r2    `Hence proved.

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q7.3

P.Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp1.3

P.Q.
Evaluate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.5

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A.

Other Exercises

Question 1.
In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that:
(i) ΔAPC and ΔBPD are similar.
(ii) If BD = 2.4 cm AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q1.1
Solution:
Two line segments AB and CD intersect each other at P.
AC || BD To prove:
(i) ΔAPC ~ ΔBPD
(ii) If BD = 2.4cm, AC = 3.6cm, PD = 4.0 cm and PB = 3.2, find length of PA and PC
Proof:
(i) In ΔAPC and ΔAPD
∠APC = ∠BPD (Vertically opp. angles)
∠PAC = ∠PBD (Alternate angles)
ΔAPC ~ ΔBPD (AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q1.2

Question 2.
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:
(i) ΔAPB is similar to ΔCPD.
(ii) PA x PD = PB x PC.
Solution:
In trapezium ABCD AB || DC
Diagonals AC and BD intersect each other at P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q2.1
To prove:
(i) ΔAPB ~ ΔCPD.
(ii) PA x PD= PB x PC.
Proof: In ΔAPB and ΔCPD
∠APB = ∠CPD (Vertically opposite angles)
∠PAB = ∠PCD (Alternate angles)
ΔAPB ~ ΔCPD (AA axiom)
\(\frac { PA }{ PC }\) = \(\frac { PB }{ PD }\)
=> PA x PD = PB x PC
Hence proved.

Question 3.
P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:
(i) DP : PL = DC : BL.
(ii) DL : DP = AL : DC.
Solution:
P is a point on side BC of a parallelogram ABCD.
DP is produced to meet AB produced at L.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q3.1
To prove:
(i) DP : PL = DC : BL
(ii) DL : DP = AL : DC.
Proof:
(i) In ΔBPL and ΔCPD
∠BPL = ∠CPD (Vertically opposite angles)
∠PBL = ∠PCD (Alternate angles)
ΔBPL ~ ΔCPD (AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q3.2

Question 4.
In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that:
(i) ΔAOB is similar to ΔCOD.
(ii) OA x OD = OB x OC.
Solution:
Given : In quadrilateral ABCD, diagonals AC and BD intersect each other at O.
AO = 2CO, BO = 2DO.
To prove:
(i) ΔAOB is similar to ΔCOD.
(ii) OA x OD = OB x OC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q4.2

Question 5.
In ΔABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that:
(i) CB : BA = CP : PA
(ii) AB x BC = BP x CA
Solution:
In ΔABC,
∠ABC = 2∠ACB
Bisector of ∠ABC meets AC in P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q5.1
To prove:
(i) CB : BA = CP : PA
(ii) AB x BC = BP x CA
Proof:
(i) In ΔABC,
BP is the bisector of ∠ABC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q5.2

Question 6.
In ΔABC; BM ⊥ AC and CN ⊥ AB; show that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q6.1
Solution:
In ΔABC,
BM ⊥ AC and CN ⊥ AB
To prove:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q6.3

Question 7.
In the given figure, DE // BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
(i) Write all possible pairs of similar triangles.
(ii) Find lengths of ME and DM.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q7.1
Solution:
In the given figure,
DE || BC
AE = 15 cm, EC = 9 cm NC = 6 cm and BN = 24 cm
(i) Write all the possible pairs of similar triangles.
(ii) Find lengths of ME and DM
Proof:
(i) In ΔABC
DE || BC
Pairs of similar triangles are
(a) ΔADE ~ ΔABC
(b) ΔADM ~ ΔABN
(c) ΔAME ~ ΔANC
(ii) ΔAME ~ ΔANC
and ΔADM ~ ΔABN
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q7.2

Question 8.
In the given figure, AD = AE and AD² = BD x EC
Prove that: triangles ABD and CAE are similar.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q8.1
Solution:
In the given figure,
AD = AE
AD² = BD x EC
To prove: ΔABD ~ ΔCAE
Proof: In ΔADC, AD = AE
∠ADE = ∠AED (Angles opposite to equal sides)
But ∠ADE + ∠ADB = ∠AED + ∠AEC = 180°
∠ADB = ∠AEC
AD² = BD x EC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q8.2

Question 9.
In the given figure, AB // DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q9.1
Solution:
In the given figure, AB || DC,
BO = 6 cm, DQ = 8 cm
Find BP x DO
In ΔODQ and ΔOPB
∠DOQ = ∠POB (Vertically opposite angles)
∠DQO = ∠OPB (Alternate angles)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q9.2

Question 10.
Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.
Solution:
In ΔABC, ∠ABC is an obtused angle,
AB =AC
P is a point on BC such that PC = 12 cm
PQ and PR are perpendiculars to the sides AB and AC respectively.
PQ = 15 cm and PR = 9 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q10.1

Question 11.
State, true or false :
(i) Two similar polygons are necessarily congruent
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
Solution:
(i) False,
(ii) True,
(iii) True,
(iv) False,
(v) True,
(vi) True,
(vii) True.

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q12.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q12.2

Question 13.
D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that CA² = CB x CD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q13.1

Question 14.
In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove ΔABC ~ ΔAMP
(ii) Find AB and BC.
Solution:
(i) In ΔABC and ΔAMP,
∠A = ∠A (Common)
∠ABC = ∠AMP (Each = 90°)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q14.1
From right triangle ABC, we have
AC² = AB² + BC² (Pythagoras Theorem)
⇒ 10² = AB² + 8²
⇒ 100 = AB² + 64
⇒ AB² = 100 – 64 = 36
⇒ AB = 6 cm
Hence, AB = 6 cm, BC = 8 cm

Question 15.
Given : RS and PT are altitudes of ΔPQR prove that:
(i) ΔPQT ~ ΔQRS,
(ii) PQ x QS = RQ x QT.
Solution:
Proof: In ΔPQT and ΔQRS,
∠PTQ = ∠RSQ (Each = 90°)
∠Q = ∠Q (Common)
ΔPQT ~ ΔQRS (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q15.1

Question 16.
Given : ABCD is a rhombus, DPR and CBR are straight lines.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q16.1
Prove that: DP x CR = DC x PR.
Solution:
Proof: In ΔAPD and ΔPRC
∠DPA = ∠CPR (Vertically opposite angles)
∠PAD = ∠PCR (Alternate angles)
ΔAPD ~ ΔPRC (AA Postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q16.2

Question 17.
Given: FB = FD, AE ⊥ FD and FC ⊥ AD.
Prove : \(\frac { FB }{ AD }\) = \(\frac { BC }{ ED }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q17.1

Question 18.
In ΔPQR, ∠Q = 90° and QM is perpendicular to PR, Prove that:
(i) PQ² = PM x PR
(ii) QR² = PR x MR
(iii) PQ² + QR² = PR²
Solution:
Given: In ΔPQR, ∠Q =90°
QM ⊥ PR.
To Prove:
(i) PQ2 = PM x PR
(ii) QR2 = PR x MR
(iii) PQ2 + QR2 = PR2
Proof: In ΔPQM and ΔPQR,
∠QMP = ∠PQR (each = 90°)
∠P = ∠P (Common)
ΔPQM ~ ΔPQR (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q18.2

Question 19.
In ΔABC, ∠B = 90° and BD x AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm, AB = 7 cm; find AD.
Solution:
In ΔABC, ∠B = 90°
∠A + ∠C = 90° …….(i)
and in ΔBDC, ∠D = 90°
∠CBD + ∠C = 90° ….(ii)
From (i) and (ii)
∠A + ∠C = ∠CBD + ∠C
∠A = ∠CBD
Similarly ∠C = ∠ABD
Now in ΔABD and ΔCBD,
∠A = ∠CBD and ∠ABD = ∠C
ΔABD ~ ΔCBD (AA Postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q19.3

Question 20.
In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q20.1
Find the lengths of PN and RM. [1997]
Solution:
In ΔLNP and ΔRLQ
∠LNP = ∠LQR (Alternate angles)
∠NLP = ∠QLR (Vertically opposite angles)
ΔLNP ~ ΔRLQ (AA Postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q20.2

Question 21.
In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that AE : EC = BE : ED. Show that ABCD is a parallelogram.
Solution:
Given : In quadrilateral ABCD, diagonals AC and BD intersect each other at E and
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q21.1
∠AEB = ∠CED (Vertically opposite angles)
ΔAEB ~ ΔCED (SAS axiom)
∠EAB = ∠ECB
∠EBA = ∠CDE
But, these are pairs of alternate angles
AB || CD …..(i)
Similarly we can prove that
AD || BC …..(ii)
from (i) and (ii)
ABCD is a parallelogram.

Question 22.
In ΔABC, AD is perpendicular to side BC and AD² = BD x DC. Show that angle BAC = 90°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q22.1
Solution:
Given: In ΔABC, AD x BC and AD² = BD x DC
To Prove: ∠BAC = 90°
Proof:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q22.2

Question 23.
In the given figure AB || EF || DC; AB = 67.5 cm. DC = 40.5 cm and AE = 52.5 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q23.1
(i) Name the three pairs of similar triangles.
(ii) Find the lengths of EC and EF.
Solution:
(i) In the figure AB || EF || DC
There are three pairs of similar triangles.
(i) ΔAEB ~ ΔDEC
(ii) ΔABC ~ ΔEEC
(iii) ΔBCD ~ ΔEBF
(ii) ΔAEB ~ ΔDEC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q23.2

Question 24.
In the given figure, QR is parallel to AB and DR is parallel to QB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q24.1
Prove that PQ² = PD x PA.
Solution:
Given: In the figure QR || AB mid DR || QB.
To Prove: PQ² = PD x PA
Proof— In ΔPQB,
DR || QB (given)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q24.2

Question 25.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E.
Prove that : EL = 2 BL.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q25.1
Given: In ||gm. ABCD, M is the mid-point A of CD.
AC is the diagonal.
BM is joined and produced meeting AD produced in E and, intersecting AC in L.
To Prove: EL = 2 BL.
Proof: In ΔEDM, and ΔMBC,
DM = MC (M is mid-point of DC)
∠EMD = ∠CMD (vertically opposite angles)
∠EDM = ∠MCB (Alternate angles)
ΔEDM = ΔMBC (ASA postulate of congruency)
ED = CB = AD (c. p. c. t.)
EA = 2 AD = 2 BC
AB = BC (opposite sides of II gm)
∠DEM = ∠MBC (c. p. c. t.)
Now in ΔELA and ΔBLC,
∠ELA = ∠BLC (vertically opposite angles)
∠DEM or ∠AEL = ∠LBC (proved)
ΔELA ~ ΔBLC (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q25.2

Question 26.
In the figure given below P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q26.1
(i) Calculate the ratio PQ : AC, giving reason for your answer.
(ii) In triangle ARC, ∠ARC = 90° and in triangle PQS, ∠PSQ = 90°.
Given QS = 6 cm, calculate the length of AR. [1999]
Solution:
Given: In ΔABC, P is a point on AB such that AP : PB = 4 : 3
and PQ || AC is drawn meeting BC in Q.
CP is joined and QS ⊥ CP and AR ⊥ CP
To Find:
(i) Calculate the ratio between PQ : AC giving reason.
(ii) In ΔARC ∠ARC= 90°
and In ΔPQS, ∠PSQ = 90°, if QS = 6 cm, calculate AR.
proof:
(i) In ΔABC, PQ || AC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q26.2

Question 27.
In the right angled triangle QPR, PM is an altitude.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q27.1
Given that QR = 8 cm and MQ = 3.5 cm. Calculate, the value of PR.
Given: In right angled ΔQPR, ∠P = 90° PM ⊥ QR, QR = 8 cm, MQ = 3.5 cm. Calculate PR [2000]
Solution:
In ΔPQM and ΔQPR,
∠PMQ = ∠QPR (each = 90°)
∠Q = ∠Q (common)
ΔPQM ~ ΔQPR (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q27.2

Question 28.
In the figure given below, the medians BD and CE of a triangle ABC meet at G.
Prove that
(i) ΔEGD ~ ΔCGB
(ii) BG = 2 GD from (i) above.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q28.1
Solution:
Given: In ΔABC, BD and CE are the medians of sides AC and AB respectively which intersect each at G.
To Prove:
(i) ΔEGD ~ ΔCGB
(ii) BG = 2 GD.
Proof: D and E are the mid points of AC and AB respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q28.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A.

Other Exercises

Prove the following Identities :
Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q1.2

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q2.2

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q3.2

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q4.2

Question 5.
sin4A – cos4 A = 2 sin2A-1 
Solution:
L.H.S. = sin4 A – cos4A = (sin2A)2-(cos2A)2
= (sin2A + cos2A) (sin2A – cos2A)     [(a2 – b2 = (a + b) (a – b)]
= 1 (sin2 A – cos2A) [∵ sin2A + cos2A = 1]
= sin2 A – (1- sin2A) (∵ cos2A = 1 – sin2A)
= sin2 A – 1 + sin2 A
= 2 sin2A-1 = R.H.S.

Question 6.
(1 – tan A)2 + (1 + tanA)2 = 2sec2A
Solution:
LHS = (1 -tanA)2 + (1 +tanA)2
= 1 + tan2 A- 2 tan A + 1 + tan2 A + 2 tanA
= 2 + 2 tan2 A = 2 (1+tan2A)
= 2 sec2A (∵ l+tan2A=sec2A)
= R.H.S.

Question 7.
Cosec4 A – cosec2 A = cot4 A + cot2 A
Solution:
L.H.S. = cosec4 A -cosec2 A
= (cosec2A)2 – cosec2A
= (1 + cot2A)2 – (1 + cot2A)
= 1 + cot4 A + 2 cot2A – 1- cot2A
= cot4 A + cot2 A = R.H.S.

Question 8.
sec A (1-sin A) (sec A + tan A) = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q8.1

Question 9.
cosec A (1 + cos A) (cosec A – cot A) = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q9.1

Question 10.
sec2 A + cosec2A = sec2 A cosec2 A
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q10.1

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q11.2

Question 12.
tan2A – sin2A = tan2 A. sin2 A
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q12.1

Question 13.
cot2 A – cos2 A = cos2 A. cot2 A
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q13.1

Question 14.
(cosecA + sinA) (cosec A – sinA) = cot2 A + cos2A
Solution:
L.H.S. = (cosec A + sin A) (cosec A – sin A)
= (cosec2A – sin2 A) [∵ (a + b) (a – b) = a2– b2]
= 1 + cot2 A – sin2 A = cot2 A + 1 – sin2A
= cot2 A + cos2 A (∵ 1-sin2A = cos2 A)
= R.H.S.

Question 15.
(sec A – cosA) (sec A + cosA) = sin2 A + tan2A
Solution:
L.H.S. = (sec A-cos A) (sec A + cos A)
= sec2 A – cos2 A
= 1 + tan2A-cos2 A
= 1-cos2 A + tan2 A
= sin2 A + tan2 A  (∵ 1- cos2A=sin2A)
= R.H.S.

Question 16.
(cos A + sin A)2 + (cos A – sin A)2 = 2
Solution:
LHS = (cos A + sin A)2 + (cos A – sin A)2
= cos2 A + sin2 A + 2 cos A sin A + cos2 A + sin2 A – 2 cos A sin A
= 2 sin2 A + 2 cos2 A
= 2 (sin2A+cos2A)
= 2 x 1=2 = R.H.S. (∵ sin2A + cos2 A = 1)

Question 17.
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q17.1

Question 18.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q18.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q18.2

Question 19.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q19.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q19.2

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q20.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q20.2

Question 21.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Solution:
L.H.S. = (sin A + cosecA)2 + (cosA+ secA)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
= sin2 A+cosec2 A+2 sin A x \(\frac { 1 }{ sinA }\) + cos2 A+sec2A + 2cosA x \(\frac { 1 }{ cosA }\)
= sin2A + cos2 A + cosec2 A + sec2A+ 2 + 2   (∵ sin2 A + cos2A= 1)
= 1 +cosec2A + sec2A + 4
= (1 + cot2 A) + (1 + tan2 A) + 5 [∵ cosec2A = 1 + cot2 A and sec2 A = 1 + tan2A]
= 1 + cot2 A + 1 + tan2 A + 5
= 7 + tan2A + cot2A = R.H.S.

Question 22.
sec2A. cosec2A = tan2A + cot2A + 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q22.1

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q23.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q23.2

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q24.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q24.2

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q25.2

Question 26.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q26.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q26.2

Question 27.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q27.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q27.2

Question 28.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q28.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q28.2

Question 29.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q29.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q29.2

Question 30.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q30.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q30.2

Question 31.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q31.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q31.2

Question 32.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q32.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q32.2

Question 33.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q33.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q33.2

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q34.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q34.2

Question 35.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q35.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q35.2

Question 36.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q36.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q36.2

Question 37.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q37.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q37.2

Question 38.
(1 +cot A-cosec A) (1 + tan A + sec A) = 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q38.1

Question 39.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q39.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q39.2

Question 40.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q40.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q40.2

Question 41.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q41.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q41.2

Question 42.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q42.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q42.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q42.3

Question 43.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q43.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q43.2

Question 44.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q44.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q44.2

Question 45.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q45.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q45.2

Question 46.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q46.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q46.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q46.3

Question 47.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q47.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q47.2

Question 48.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q48.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q48.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Other Exercises

Question 1.
Find AD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q1.3

Question 2.
In the following diagram.
AB is a floor-board. PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q2.3

Question 3.
Calculate BC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q3.2

Question 4.
Calculate AB .
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q4.2

Question 5.
The radius of a circle is given as 15cm and chord AB subtends an angle of 131° at the centre C of the circle. Using trigonometry, Calculate :
(i) the length of AB;
(ii) the distance of AB from the centre C.
Solution:
Chord AD substends an angle of 131° at the centre. Join CA, CB and draw CD ⊥ AB which bisects AB at D.
(In ∆CAB)
∵ CA = CB (radii of the same circle)
∴ ∠CAB = ∠CBA
But ∠CAB + ∠CBA = 180°- 131° = 49°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q5.2

Question 6.
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \(\frac { 5 }{ 12 }\). On walking 192 metres towards the tower; the tangent of the angle is found to be \(\frac { 3 }{ 4 }\). Find the height of the tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q6.3

Question 7.
A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is a and that of the top of flagstaff is β. Prove that the height of the tower is :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.4

Question 8.
With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m.
The man’s eye is 2 m above the ground. He observes the angle of elevation of C. The top of the pole, as x°, where tan x° = \(\frac { 2 }{ 5 }\) . calculate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q8.1
(i) the distance AB in m;
(ii) the angle of elevation of the top of the pole
when he is standing 15 m from the pole. Give your answer to the nearest degree. [1999]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q8.3

Question 9.
The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same striaght line with it are complementary.Prove that height of the tower is \( \sqrt{ab} \) metre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q9.2

Question 10.
From a window A. 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x = \(\frac { 5 }{ 2 }\) and the angle of depression of the foot D of the tower is y°, where tany° = \(\frac { 1 }{ 4 }\).(See the figure given below). Calculate the height CD of the tower in metres. [2000]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q10.1
Solution:
Let CD be the height of the tower and height of window A from the ground = 10m
In right ∆AEC,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q10.3

Question 11.
A vertical tower is 20 m high, A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ? [2001]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q11.2

Question 12.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate :
(i) the width of the river and
(ii) the height of the tree.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q12.1
Let TR be the tree of height x m and y be the width AR of the river,
then ∠B = 30° and A = ∠60° , AB = 50 m.
Now in right ∆ATR,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q12.3

Question 13.
A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is 60° and the angle of elevation of the top of the pole as seen from the foot of the tower is 30°. Find
(i) the height of the tower.
(ii) the horizontal distance between the pole and the tower.
Solution:
Let PQ is the pole and TS is the tower. PQ = 20 m.
Let TS = h and QS = x Angles of elevation from Q to T A T is 60° and from S to P is 30°.
In the ∆PQS
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q13.2

Question 14.
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°.
Find : (i) the height of the tower, if the height of the pole is 20m;
(ii) the height of the pole, if the height of the tower is 75 m.
Solution:
Let PQ is the pole and TW is the tower
Angle of elevation from T to P is 60° and angle of depression from P to W is 30°
∴ ∠PWQ = 30° = ∠RPW ( ∠ Altanate angles)
(i) In first case when height of pole OQ = 20m, Then in right ∆ PQW
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q14.2

Question 15.
From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.
Solution:
Let AQ is the sea-level
P is a point 36 m above sea-level
∴ PQ = 36
Let B be the bird and R is its reflection in the water and angle of elevation of the bird B at P is 30° and angle of depression of the reflection of the bird at R is 60°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q15.2

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q16.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q16.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q16.3

Question 17.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of the light house in horizontal line with its base , are 30° and 40° respectively . Find the distance between the two ships.Give your answer correct to the nearest meter. [2012]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q17.2

Question 18.
In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.1
(i) the horizontal distance between AB and CD.
(it) the height of the lamp post.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.4

Question 19.
An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. (2014)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q19.2

Question 20.
The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24° respectively. Find the height of the two towers. Give your answer correct to 3 significant figures. (2015)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q20.1
Solution:
AB and CD are two towers which are 120 m apart
i.e. BD= 120m
Angles of elevation of the top and angle of depression of bottom of the first tower observed from the top of second tower is 30° and 24°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q20.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q20.3

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E.

Other Exercises

Question 1.
Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio, 3 : 5. Find its co-ordinates of point P. Also, find the equation of the line through P and parallel to 3x + 5y = 7
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q1.2

Question 2.
The line segment joining the points A (3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.
Solution:
Point P, divides the line segment A (3, -4) and B(-2, 1) in the ratio of 1 : 3
Let co-ordinates of P be (x, y), then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q2.2

Question 3.
A line 5x + 3y + 15 = 0 meets y -axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution:
P lies on y-axis and let the co-ordinates of P be (0, y)
P lies also on the line 5x + 3y + 15 = 0 it will satisfy it.
5 x 0 + 3y + 15 = 0
⇒ 3y = -15
⇒ y = -5
Co-ordinates of P are (0, -5)
Now, writing the line x – 3y + 4 = 0 is form of y = mx + c
-3y = -x – 4
⇒ 3y = x + 4
⇒ y = \(\frac { 1 }{ 3 }\) x + \(\frac { 4 }{ 3 }\)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q3.1

Question 4.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other. [2003]
Solution:
Writing, the line kx – 5y + 4 = 0 in form of y = mx + c
⇒ -5y = -kx – 4
⇒ 5y = kx + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q4.1

Question 5.
A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid point of the segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of A and B.
(iii) the co-ordinates of M. (2003)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q5.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q5.3
(ii) Let. co-ordinates of A be (x, 0) and of B be (0, y) which lie On the line.
Substituting, the co-ordinates in (i)
x + 0 = 3 ⇒x = 3
Co-ordinates of A are (3, 0)
Again 0 + y = 3 ⇒ y = 3
Co-ordinates of B are (0,3)
(iii) M is the mid-point of AB.
Co-ordinates of M wil be (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

Question 6.
(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Solution:
Co-ordinates of A and C of rhombus ABCD are (1, 5) and (-3, -1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q6.2
⇒ 3y – 6 = -2x – 2
⇒ 2x + 3y = 6 – 2
⇒ 2x + 3y = 4

Question 7.
Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution:
Three vertices of a square ABCD are A (3, 2), B (6, -2) and C (2, -5)
Let, co-ordinates of fourth vertex D be (x, y)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.4

Question 8.
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
Solution:
Slope of line x = 3y + 2 or 3y = x – 2 ….(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q8.3

Question 9.
A straight line passes through the point (3, 2) and the portion of this line intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q9.1
Let, the line intersects x-axis at A and y-axis at B.
Let, co-ordinates of A (x, o) and of (o, y)
But (3, 2) is the mid-point of AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q9.2

Question 10.
Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x -2y = 1.
Solution:
7x + 6y = 71 ….(i)
5x – 8y = -23 ….(ii)
Multiply (i) by 4 and (ii) by 3,
28x + 24y = 284
15x – 24y = -69
On adding (i) and (ii), we get:
43x = 215
x = 5
Substituting, the value of x in (i)
7 x 5 + 6y = 71
35 + 6y = 71
⇒ 6y = 71 – 35 = 36
⇒ y = 6
Point of intersection of these lines is (5, 6)
Now slope of line 4x – 2y = 1
⇒ 4x – 1 = 2y
⇒ y = 2x – \(\frac { 1 }{ 2 }\) is 2
Slope of line through the point of intersection and perpendicular to 4x – 2y = 1 is \(\frac { -1 }{ 2 }\)
Equation of the line y – y1 = m (x – x1)
⇒ y – 6 = \(\frac { -1 }{ 2 }\) (x – 5)
⇒ 2y – 12 = -x + 5
⇒ x + 2y = 5 + 12 = 17
⇒ x + 2y = 17

Question 11.
Find the equation of the line which is perpendicular to the line \(\frac { x }{ a }\) – \(\frac { y }{ b }\) = 1 at the point where this line meets y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q11.2

Question 12.
O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of ∆OAB through vertex O.
(ii) the equation of altitude of ∆OAB through vertex B.
Solution:
(i) Let, mid-point of AB be D.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q12.2

Question 13.
Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does line 3x = y + 1 bisect the line segment joining the two given points ?
Solution:
Slope of the line joining the points (-2, 3) and (4, 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q13.1
Yes, these are perpendicular to each other
Let P be the mid-point of the line joining the points (-2, 3) and (4, 1)
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q13.2
This point (1, 2) satisfies the equaion 3x = y + 1 then, it will bisect the line joining the given point
now, substituting the value of x and y. in 3x = y + 1
⇒ 3 x 1 = 2 + 1
⇒ 3 = 3. which is true.
Yes, the line 3x = y + 1 is the bisector.

Question 14.
Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).
Solution:
Equation of the given line is x cos 30° + y sin 30° = 2
y sin 36° = -x cos 30° + 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q14.1

Question 15.
Find the value of k such that the line (k – 2) x + (k + 3) y – 5 = 0 is :
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Solution:
Writing the given equation in the form of y = mx + c
(k – 2) x + (k + 3) y – 5 = 0 ….(i)
⇒ (k + 3) y = – (k – 2) x + 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q15.2

Question 16.
The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7) write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i) meets BC.
Solution:
Vertices of ∆ABC are A (0, 5),B (-1, -2) and C (11, 7)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q16.1
(ii) Let, the line through A meets BC in P
P is point of intersection of these two lines.
3x – 4y = 5 ……… (i)
4x + 3y = 15 …….. (ii)
On solving (i), (ii) we get
x = 3, y = 1
Co-ordinates of Pare (3, 1)

Question 17.
From the given figure, find :
(i) the co-ordinates of A, B and C.
(ii) the equation of the line through A and parallel to BC. (2005)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q17.1
Solution:
(i) From the figure, we see that co-ordinates of A are (2, 3), of B are (-1, 2) of C and (3, 0)
(ii) Slope of line BC is (m)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q17.2
⇒ x + 2y – 6 – 2 = 0
⇒ x + 2y – 8 = 0
⇒ x + 2y = 8

Question 18.
P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R. (2004)
Solution:
Let (x, y) be the co-ordinates of M, the mid-point of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q18.2
⇒ x + 2y – 8 = 0
⇒ x + 2y = 8

Question 19.
A (8, -6), B (-4, 2) and C (0, -10) are the vertjces of a triangle ABC. If P is the mid-point of AR and Q is the mid-point of AC, use co ordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ.
Solution:
In ∆ABC, co-ordinates of A, B and C are (8, -6), (-4, 2) and C (0, -10) respectively.
P and Q are the mid-points of AB and AC respectively
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q19.2
Slopes of PQ and BC are same.
These are parallel to each other.
Quad. PBCQ is trapezium

Question 20.
A line AB meets the x-axis at point A and y-axis at point B. The point P(-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find :
(i) the co-ordinates of A and B.
(ii) equation of line through P, and perpendicular to AB.
Solution:
Line AB intersects x-axis at A and y-axis at B.
(i) Let co-ordinates of A be (x, 0) and of B be (0, y)
Point P (-4, -2) intersects AB in the ratio 1 : 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q20.2

Question 21.
A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from die positive side of y-axis. Find the equation of the line. (1992)
Solution:
Let line intersects x-axis at P (-2, 0) and cuts off an intercept of 3 units at Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q21.1

Question 22.
Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. (2002)
Solution:
x-intercept = 4
Co-ordinates of that point = (4, 0)
The co-ordinates of the given point (2, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q22.1

Question 23.
The given figure (not drawn to scale) shows two straight lines AB and CD. If’ equation of the line AB is : y = x + 1 and equation of line CD is : y = √3 x – 1. Write down the inclination of lines AB and CD; also, find the angle 6 between AB and CD. (1989)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q23.1
Solution:
Equation of line AB is y = x + 1
and equation of line CD is y = √3 x – 1
Slope of AB = 1
tanθ = 1
⇒ θ = 45°
Inclination angles of AB = 45°
Slope of CD = tanθ = √3 = tan 60°
⇒ θ = 60°
Inclination angle of CD = 60°
In ΔPQR,
Ext. ∠RQX = ∠RPQ + ∠PRQ (Exterior angles is equal to sum of its interior opposite angles)
⇒ 60° = 45° + θ
⇒ θ = 60° – 45° = 15°
⇒ θ = 15°

Question 24.
Write down the equation of the line whose gradient is \(\frac { 3 }{ 2 }\) and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3. (1996, 2001)
Solution:
P divides the line segment AB in which A (-2, 6) and B (3, -4) in the ratio 2 : 3
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q24.1

Question 25.
The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.
Solution:
Let points are A (6, 4) and B (7, -5)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q25.1

Question 26.
Points A and B have coordinates (7, -3) and (1, 9) respectively. Find
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ of (-2, p) lies on it.
Solution:
Coordinates of A are (7, -3), of B = ( 1, 9)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q26.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q26.2

Question 27.
A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the mid point of AB. Find the
(i) Coordinates of A and B.
(ii) Slope of line AB.
(iii) equation of line AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q27.1
Solution:
As P (2, -3) is mid-point of AB.
Let coordinates of B be (0, y) and coordinates of A be (x, 0)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q27.2

Question 28.
The equation of a line is 3x + 4y – 7 = 0. Find:
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
Given line 3x + 4y -7 = 0
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q28.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q28.2
⇒ 3 (y – 4) = 4 (x – 2)
⇒ 3y – 12 = 4x – 8
⇒ 4x – 3y – 8 + 12 = 0
⇒ 4x – 3y + 4 = 0

Question 29.
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4). Find :
(i) co-ordinates of A
(ii) equation of diagonal BD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q29.1
(i) In || gm ABCD, A (x, y), B (5, 8), C (4, 7) and D (2, -4)
The diagonals of ||gm bisect each other
O is said point of AC and BD
Now if O is mid point of BD then its co-ordinates will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q29.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q29.3
⇒ y + 4 = 4x – 8
⇒ 4x – y -8 – 4 = 0
⇒ 4x – y – 12 = 0 or 4x – y = 12

Question 30.
Given equation of line L1 is y = 4.
(i) Write the slope of line L2 if L2 is the bisector of angle O.
(ii) Write the co-ordinates of point P.
(iii) Find the equation of L2.
Solution:
(i) Equation of line L1 is y = 4
L2 is the bisector of ∠O
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q30.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q30.2

Question 31.
Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q31.1
(i) equation of AB
(ii) equation of CD
Solution:
Co-ordinates of A and B are (-5, 4) and (3, 3) respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q31.2

Question 32.
Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1
Solution:
x-intercept of the line = -3
and is perpendicular to the line
3x + 5y = 1
5y = 1 – 3x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q32.1

Question 33.
A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find :
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M.
Solution:
A line passing through the two points P (-1, 4) and Q (5, -2) intersects x-axis at point A and y- axis at point B.
M is mid-point of AB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q33.1
(i) Equation of line AB will be
y – y1 = m (x – x1)
⇒ y – 4 = -1 (x + 1)
⇒ y – 4 = – x – 1
⇒ y + x = -1 + 4
⇒ x + y = 3
(ii) The line intersect x-axis at A and OA = 3 units
Co-ordinates of A are (3, 0) and the line intersects y-axis at B and OB = 3 units
Co-ordinates of B are (0, 3)
(iii) M is mid-point of AB
Co-ordinates of M are (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

Question 34.
In the given figure, line AB meets y-axis at point A. Line through C (2, 10) and D intersects line AB at right angle at point P. Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q34.1
(i) equation of line AB.
(ii) equation of line CD.
(iii) co-ordinates of points E and D.
Solution:
In the given figure, AB meets y-axis at point A.
Line through C (2, 10) and D intersects line AB at P at right angle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q34.2
Equation of CD
y – 10 = 3 (x – 2)
⇒ y – 10 = 3x – 6
⇒ 3x – y + 10 – 6 = 0
⇒ 3x – y + 4 = 0
(iii) Co-ordinates of D which is on x-axis
3x – y + 4 = 0
3x – 0 + 4 = 0
⇒ 3x + 4 = 0
⇒ 3x = -4
⇒ x = \(\frac { -4 }{ 3 }\)
Co-ordinates of D are (\(\frac { -4 }{ 3 }\) , 0)
E is also on x-axis
x + 3y = 18
Substituting, y = 0, then
x + 0= 18
⇒ x = 18
Co-ordinates of E are (18, 6)

Question 35.
A line through point P (4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.
Solution:
A line through P (4, 3) meets x-axis at A and the y-axis at B. If BP is double of PA.
Draw BC || x-axis
and PC || y-axis
In ∆PAD and ∆PBC
∠P = ∠P (common)
∠D = ∠C (each 90°)
∆PAD ~ ∆PBC
PB = 2PA ⇒ PA = \(\frac { 1 }{ 2 }\) PB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q35.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q35.2
2y – 6 = 3x – 12
⇒ 3x – 2y – 12 + 6 = 0
⇒ 3x – 2y – 6 = 0
⇒ 3x – 2y = 6

Question 36.
Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.
Solution:
Equation of given two intersecting lines are 2x – y = 1 and 3x + 2y = -9 Which make an angle of 30°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q36.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q36.2

Question 37.
Find the equation of the line through the points A (-1, 3) and B (0, 2). Hence, show that the points A, B and C (1, 1) are collinear.
Solution:
The given points are A (-1, 3) and B (0, 2) and co-ordinates of a point C are (1, 1)
Now slope of the line joining A and B
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q37.1
Equation of the line y – y1 = m(x – x1)
⇒ y – 2 = -1 (x – 0)
⇒ y – 2 = -x
⇒ x + y – 2 = 0
Point C (1, 1) will be on AB if it satisfy
1 + 1 – 2 = 0
⇒ 0 = 0
Point C lies on AB
Hence points A, C and B are collinear.

Question 38.
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2), find:
(i) the co-ordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)
Solution:
Three vertices of a ||gm ABCD taken an order are A (3, 6), B (5, 10) and C (3, 2)
Join diagonals AC and BD which bisect each other at O.
O is mid-point of AC as well as of BD
Now co-ordinates of O will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q38.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q38.2

Question 39.
In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q39.1
(i) Write the co-ordinates of A.
(ii) Find the length of AB and AC.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC. (2015)
Solution:
In the given figure,
ABC is a triangle and BC || y-axis
AB and AC intersect the y-axis at P and Q respectively.
(i) Co-ordinates of A are (4,0).
(ii) Length of AB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q39.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q39.3

Question 40.
The slope of a line joining P (6, k) and Q (1 – 3k, 3) is \(\frac { 1 }{ 2 }\). Find :
(i) k
(ii) mid-point of PQ, using the value of ‘A’ found in (i). (2016)
Solution:
(i) Slope of the line joining P(6, k) and Q (1 –
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q40.1

Question 41.
A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1 : 2.
(i) Find the coordinates of A and B.
(ii) Find the equation of the line through P and perpendicular to AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q41.1
Solution:
(i) Since, A lies on the x-axis,
let the coordinates of A be (x, 0).
Since B lies on the y-axis,
let the coordinates of B be (0, y).
Let m = 1 and n = 2.
Using section formula,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q41.2

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D.

Other Exercises

Question 1.
Find the slope and y-intercept of the line :
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5
Solution:
(i) y = 4 ⇒ y = 0x + 4
Here slope = 0 and y-intercept = 4
(ii) ax – by = 0
⇒ by = ax
⇒ y = \(\frac { a }{ b }\) x + 0
Here, slope = \(\frac { a }{ b }\) and y-intercept = 0
(iii) 3x – 4y = 5
⇒ – 4y = 5 – 3x
⇒ 4y = 3x – 5
⇒ y = \(\frac { 3 }{ 4 }\) x + \(\frac { -5 }{ 4 }\)
Here, slope = \(\frac { 3 }{ 4 }\) and y- intercept = \(\frac { -5 }{ 4 }\)

Question 2.
The equation of a line is x – y = 4. Find its – slope and y-intercept. Also, find its inclination.
Solution:
x – y = 4
writing the equation in form of y = mx + c
x = 4 + y
⇒ y = x – 4
Slope = 1 and y-intercept = – 4
Slope = 1
⇒ tanθ = 1
⇒ θ = 45°

Question 3.
(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0 ?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7 ?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1 ?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Solution:
(i) 3x + 4y + 7 = 0
Writing the equation in form of y = mx + c
4y = -3x – 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.4

Question 4.
Find the slope of the line which is parallel to:
(i) x + 2y + 3 = 0
(ii) \(\frac { x }{ 2 }\) – \(\frac { y }{ 3 }\) – 1 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q4.2

Question 5.
Find the slope of the line which is perpendicular to:
(i) x – \(\frac { y }{ 2 }\) + 3 = 0
(ii) \(\frac { x }{ 3 }\) – 2y = 4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q5.2

Question 6.
(i) Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x- ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
Solution:
(i) Writing the given equations in the form of y = mx + c, we get:
-by = -2x -5
by = 2x + 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q6.2

Question 7.
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Solution:
Writing the given equations 2x – y + 5 = 0 and px + 3y = 4 in form of y = mx + c
2x – y + 5 = 0
-y = -2x -5
y = 2x + 5
Here, slope of the line = 2
Again, px + 3y = 4
3y = – px + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q7.1

Question 8.
The equation of a line AB is 2x – 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Solution:
The line AB is given.2* – 2y + 3 = 0
Writing it in the form of y = mx + c
-2y = -2x – 3
⇒ 2y = 2x + 3
⇒ y = x + \(\frac { 3 }{ 2 }\)
Here, slope of the line = 1
Angle of inclination = tanθ
tanθ = 1
θ = 45°

Question 9.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
Solution:
Writing the given lines 4x + 3y = 9 and px – 6y + 3 = 0 in the form of y = mx + c
4x + 3y = 9
⇒ 3y = – 4x + 9
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q9.1

Question 10.
If the lines y = 3x + 7 and 2y +px = 3 are perpendicular to each other, find the value of p. (2006)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q10.1

Question 11.
The line through A (-2, 3) and B (4, 6) is perpendicular to the line 2x – 4y = 5. Find the value of b.
Solution:
Gradient (mx) of the line passing through the points A (-2, 3) and B (4, b)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q11.1

Question 12.
Find the equation of the line passing through (-5, 7) and parallel to
(i) x-axis
(ii) y-axis.
Solution:
(i) Slope of the line parallel to x-axis = 0
Equation of line passing through (-5, 7) whose slope is 0.
y – 7 = 0 [x – (-5)]
⇒ y – 7 = 0
⇒ y = 7
(ii) Slope of the line parallel to y-axis = 0
y – y1 = m (x – x1)
⇒ 0 = x – x1
⇒ x + 5 = 0

Question 13.
(i) Find the equation of the line passing through (5, -3) and parallel to x – 3y = 4.
(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1). (2007)
Solution:
(i) Writing the equation x – 3y = 4 in form of y = mx + c
-3y = -x + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q13.2
⇒ 2y – 2 = -3x
⇒ 3x + 2y – 2 = 0
Which is the required equation.

Question 14.
Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.
Solution:
Writing the equation 4x + 5y = 6 in form of y = mx + c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q14.1

Question 15.
Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).
Solution:
The perpendicular of the line segment bisects it.
Co-ordinates of mid-point of the line segment which is obtained by joining the points (6, -3) and (0, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q15.1
Slope of the line perpendicular to it = 1 (Product of slopes = -1)
Equation of the perpendicular bisector is y – y1 = m (x – x1)
y – 0 = 1 (x – 3)
y = x – 3

Question 16.
In the following diagram, write down:
(i) the co-ordinates of the points A, Band C.
(ii) the equation of the line through A and parallel to BC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q16.1
Solution:
(i) From the figure, the see that co-ordinates of A are (2, 3), B are (-1, 2) and C are (3, 0)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q16.2

Question 17.
B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q17.1

Question 18.
A = (7, -2) and C = (-1, – 6) are the vertices of a square ABCD. Find the equation of the diagonals AC and BD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q18.1

Question 19.
A (1, -5), B (2, 2) and C (-2, 4) are the vertices of the ∆ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.
Solution:
(i) Let D be the mid-point of BC
co-ordinates mid-point of
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q19.3

Question 20.
(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin. [1995]
Solution:
(i) Write the equation 2y = 3x + 5 in the form of y = mx + c. We get:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q20.1
y – 2 = \(\frac { -2 }{ 3 }\) (x – 3)
⇒ 3y – 6 = -2x + 6
⇒ 2x + 3y = 6 + 6
⇒ 2x + 3y = 12 …… (i)
(ii) AB meets the x-axis at A
ordinate (y) of A = 0 i.e. y = 0
Substituting, the value of y in (i)
2x + 3 x 0 = 12
⇒ 2x = 12
⇒ x = 6
Co-ordinates of A are (6, 0)
Again. AB meets y-axis at B
Abscissa of B = 0 i.e. x = 0
Substituting the value of x in (i)
2 x 0 + 3y = 12
⇒ y = 4
Co-ordinates of B are (0, 4)
Area of ∆OAB = \(\frac { 1 }{ 2 }\) x Base x altitude
= \(\frac { 1 }{ 2 }\) x 4 x 6 = 12 square units

Question 21.
The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.
Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.
Solution:
The line 4x – 3y + 12 = 0 meets x-axis at A.
Ordinates of A = 0. i.e. y = 0
Substituting, the value of y in the equation
4x – 3 x 0 + 12 = 0
⇒ 4x + 12 = 0
⇒ 4x = -12
⇒ x = -3
Co-ordinates of A are (-3, 0)
Writing the equation 4x – 3y + 12 = 0 in form of y = mx + c
⇒ -3y = -4x – 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q21.1

Question 22.
The point P is the foot of perpendicular from A (-5, 7) to the line is 2x – 3y + 18 = 0. Determine:
(i) the equation of the line AP.
(ii) the co-ordinates of P.
Solution:
Write the equation in form of y = mx + c
2x – 3y + 18 = 0
⇒ -3y = -2x – 18
⇒ y = \(\frac { 2 }{ 3 }\) x + 6 (Dividing by 3)
Slope of the line = \(\frac { 2 }{ 3 }\)
and slope of the line perpendicular to it = \(\frac { -3 }{ 2 }\) (Product of slopes = -1)
(i) Equation of line AP perpendicular to the given line and drawn through A (-5, 7)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q22.1
y – y1 = m (x – x1)
⇒ y – 7 = \(\frac { -3 }{ 2 }\) (x + 5)
⇒ 2y – 14 = -3x – 15
⇒ 3x + 2y – 14 + 15 = 0
⇒ 3x + 2y + 1 = 0
(ii) P is the point of intersection of these lines
we will solve their equations
2x – 3y + 18 = 0 ….(i)
3x + 2y + 1 = 0 ….(ii)
Multiplying (i) by 2 and (ii) by 3, we get
4x – 6y + 36 = 0
9x + 6y + 3 = 0
Adding, we get:
13x + 39 = 0
⇒ 13x = -39
⇒ x = -3
Now, substituting the value of x in (i)
2(-3) – 3y + 18 = 0
⇒ -6 – 3y + 18 = 0
⇒ -3y + 18 – 6 = 0
⇒ -3y + 12 = 0
⇒ -3y = -12
⇒ 3y = 12
⇒ y = 4
Co-ordinates of P are (-3, 4)

Question 23.
The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC. If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q23.1
AB meets y- axis at P
abscissa of P = 0 i. e. x = 0
Substituting the value of y in (i)
0 + y = 4
⇒ y = 4
Co-ordinates of P are (0, 4)
BC meets x-axis at Q
ordinate of Q = 0 i.e. y = 0
Substituting, the value of y in (ii),
2x + 0 = 6
⇒ 2x = 6
⇒ x = 3
Co-ordinates of Q are (3, 0)

Question 24.
Match the equations A, B, C, and D with the lines L1, L2, L3 and L4, whose graphs are roughly drawn in the given diagram.
A = y = 2x;
B = y – 2x + 2 = 0;
C = 3x + 2y = 6;
D = y = 2 [1996]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q24.1
Solution:
A → L3,
B → L4,
C → L2,
D → L1

Question 25.
Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line. (2014)
Solution:
A (a, 3), B (2,1) and C (5, a) are collinear.
Slope of AB = Slope of BC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q25.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C.

Other Exercises

Question 1.
Find the equation of line whose :
(i) y-intercept = 2 and slope = 3,
(ii) y-intercept = -1 and slope = \(\frac { -3 }{ 4 }\)
Solution:
(i) The point whose y-intercept = 2 will be (0, 2) and slope (m) = 3.
Equation of line will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q1.1

Question 2.
Find the equation of a line whose :
(i) y-intercept = -1 and inclination = 45°
(ii) y-intercept = 3 and inclination = 30°
Solution:
(i) The point whose y-intercept is -1, will be (0, -1) and inclination = 45°
Slope (m) = tan 45° = 1
Equation will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q2.1

Question 3.
Find the equation of the line whose slope is \(\frac { -4 }{ 3 }\) and which passes through (-3, 4).
Solution:
Slope of the line (m) = \(\frac { -4 }{ 3 }\)
The point from which the line passes (-3, 4)
Equation of line will be y – y1 = m (x – x1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q3.1

Question 4.
Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.
Solution:
The line passes through the point (5, 4) and angle of inclination = 60°
slope (m) = tan 60° = √3
Equation of line
y – y1 = m (x – x1)
⇒ y – 4 = √3 (x – 5)
⇒ y – 4 = √3 x – 5√3
⇒ y = √3 x + 4 – 5√3

Question 5.
Find the equation of the line passing through:
(i) (0, 1) and (1, 2)
(ii) (-1, -4) and (3, 0)
(iii) (4, -2) and (5, 2)
Solution:
Two given points are (0, 1) and (1, 2)
Slope of the line passing through these two
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q5.2

Question 6.
The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :
(i) The gradient of PQ
(ii) The equation of PQ,
(iii) The co-ordinates of the point where PQ intersects the x-axis.
Solution:
Two points P (2,-6) and Q (-3, 5) are given.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q6.1
⇒ 5y – 30 = x – 2
⇒ 5y = x – 2 + 30
⇒ 5y = x + 28 ….(i)
(iii) Co-ordinates of the point where PQ intersects x-axis will be = 0
substituting, the value of y in (i)
5 x 0 = x + 28 ⇒ x + 28 = 0 ⇒ x = -28
Co-ordinates of point are (-28, 0)

Question 7.
The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find :
(i) the equation of AB
(ii) the co-ordinates of the point where the line AB intersects they- axis.
Solution:
Slope of the line passing through two points A (-3, 4) and B (2, -1) will be :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q7.1
Its abscissa = 0
substituting, the value of x = 0 in (i)
0 + y = 1
y = 1
Co-ordinates of point = (0, 1)

Question 8.
The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equations of AB and CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q8.1
Solution:
Two lines AB and CD intersect each other at P (3, 4)
AB inclined at angle of 45° and CD at angle of 60° with the x-axis.
Slope of AB = tan 45° = 1
and slope of CD = tan 60° = √3
Now, equation of line AB will be
y – y1 = m (x – x1)
⇒ y – 4 = 1 (x – 3)
⇒ y – 4 = x – 3
⇒ y = x – 3 + 4
⇒ y = x + 1
(ii) Equation of CD will be :
y – y1 = m (x – x1)
⇒ y – 4 = √3 (x – 3)
⇒ y – 4 = √3 x – 3√3
⇒ y = √3 x – 3√3 + 4
⇒ y = √3 x + 4 – 3√3

Question 9.
In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.
Solution:
AD is median
D is mid point of BC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q9.2
Equation of AD
y – y1 = m (x – x1)
⇒ y + 1 = -6 (x – 4)
⇒ y + 1 = -6x + 24
⇒ y + 6x = -1 + 24
⇒ 6x + y = 23

Question 10.
The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex, C = (7, 5). Find the equations of BC and CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q10.2
ABCD is a ||gm in which AB = CD || x-axis
∠A = 60° and C (7, 5)
(i) CD || AB || x-axis ,
Equation of CD will be
y – y1 = m (x – x1)
⇒ y – 5 = 0 (x – 7)
⇒ y – 5 = 0
⇒ y = 5
BC || AD
Slope of BC = tan 60° = √3
Equation of BC will be
y – y1 = m (x – x1)
⇒ y – 6 = √3 (x – 7)
⇒ y – 6 = √3 x – 7√3
⇒ y = √3 x + 6 – 7√3

Question 11.
Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.
Solution:
Point of intersection of two lines
x + 2y = 7 ….(i)
x – y = 4 ….(ii)
Subtracting, we get
3y = 3
y =1
Substituting, the value of y in (ii)
x – 1 = 4
⇒ x = 4 + 1 = 5
Point of intersection is (5, 1)
Slope of the line passing through origin (0, 0) and (5, 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q11.1
Equation of line will be
y – y1 = m (x – x1)
⇒ y – 5 = \(\frac { 1 }{ 2 }\) (x – 1)
⇒ 5y – 25 = x – 1
⇒ 5y = x – 1 + 25 = x + 24
⇒ 5y = x + 24

Question 12.
In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q12.2

Question 13.
A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q13.1
Slope of line through A, perpendicular to BC = -(-1) = 1
Now, the equation of line through A (0, 3) is
y – y1 = m (x – x1)
y – 3 = 1 (x – 0)
⇒ y – 3 = x
⇒ y = x + 3

Question 14.
Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).
Solution:
Slope of the line joining the points (1, 4) and (2, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q14.1
Slope of line perpendicular to the above line = 1
Equation of line passing through (-1, 2)
y – y1 = m (x – x1)
⇒ y – 2 = 1 [x -(-1)]
⇒ y – 2 = x + 1
⇒ y = x + 1 + 2
⇒ y = x + 3

Question 15.
Find the equation of the line, whose :
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
(iv) x-intercept = 3 and y-intercept = -6
Solution:
(i) When x-intercept = 5, then point will be (5, 0)
and when y-intercept = 3, then point will be (0, 3)
Slope of the line passing through these points
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q15.3

Question 16.
Find the equation of the line whose slope is \(\frac { -5 }{ 6 }\) and x-intercept is 6.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q16.1

Question 17.
Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Solution:
x-intercept of the line = 5
Point = (5, 0)
Slope of the line passing through the point (-3, 2)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q17.1

Question 18.
Find the equation of the line through (1, 3) and making an intercept of 5 on the y- axis.
Solution:
The line makes y-intercept = 5
Point = (0, 5)
Slope of the line passing through the point (1, 3) and (0, 5)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q18.1
Equation of the line
y – y1 = m (x – x1)
⇒ y – 3 = -2 (x – 1)
⇒ y – 3 = -2x + 2
⇒ 2x + y = 2 + 3
⇒ 2x + y = 5

Question 19.
Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q19.2
Solution:
(i) Slope of line AB = tan 45° = 1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = 1 (x + 2)
⇒ y = x + 2
⇒ x – y + 2 = 0
(ii) Slope of line CD = tan (-45°) = -1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = -1 (x + 2)
⇒ y = -x – 2
⇒ y + x + 2 = 0
⇒ x + y + 2 = 0

Question 20.
The line through P (5, 3) intersects y axis at Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q20.1
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.
Solution:
(i) Here θ = 45°
So, slope of the line = tanθ = tan 45° = 1
(ii) Equation of the line through P and Q is
y – 3 = 1 (x – 5)
⇒ y – x + 2 = 0
(iii) Let the coordinates of Q be (0, y)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q20.2

Question 21.
Write down the equation of the line whose gradient is \(\frac { -2 }{ 5 }\) and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3 : 1.
Solution:
Slope of the line m = \(\frac { -2 }{ 5 }\)
P divides the line AB, whose co-ordinates are (4, -8) and (12, 0) in the ratio of 3 : 1
Co-ordinates of P be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q21.1

Question 22.
A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC. Find :
(i) the co-ordinates of the centroid of ΔABC.
(ii) the equation of a line, through the centroid and parallel to AB. [2002]
Solution:
(i) Co-ordinates of vertices of ΔABC are A (1, 4), B (3, 2), C (7, 5)
and let G be the centroid of ΔABC.
Co-ordinates of G are
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q22.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q22.2

Question 23.
A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point F in AC; such that AP : CP = 2 : 3.
Solution:
P divides AC in the ratio of 2 : 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q23.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q23.2

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B.

Other Exercises

Question 1.
Find the slope of the line whose inclination is :
(i) 0°
(ii) 30°
(iii) 72° 30′
(iv) 46°
Solution:
(i) Slope of line whose inclination is 0° = tanθ = tan 0° = 0
(ii) Slope of line whose inclination is 30° = tan 30° = \(\frac { 1 }{ \surd 3 }\)
(iii) Slope of line whose inclination is 72° 30′ = tan 72°30’ = 3. 1716 (Using tables)
(iv) Slope of line whose inclination is 46° = tan 46° = 1.0355 (Using tables)

Question 2.
Find the inclination of the line whose slope is:
(i) 0
(ii) √3
(iii) 0.7646
(iv) 1.0875
Solution:
Slope of a line = tanθ. Where θ is the inclination
(i) When slope is θ. then tanθ = 0 ⇒ θ = 0°
(ii) When slope is θ, then tanθ = √3 ⇒ θ = 60°
(iii) When slope is 0.7646, then tanθ = 0.7646 ⇒ θ = 37°24′ (Using tables)
(iv) When slope is 1.0875, then tanθ = 1.0875 ⇒ θ = 47°24′ (Using tables)

Question 3.
Find the slope of the line passing through the following pairs of points :
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b ) and (b, -a)
Solution:
We know that, slope of a line which passes
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q3.2

Question 4.
Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q4.1

Question 5.
Find the slope of the line perpendicular to AB if:
(i) A = (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q5.2

Question 6.
The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.
Solution:
Slope of the line passing through two points (0, 2) and (-3, -1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q6.1

Question 7.
The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.
Solution:
Slope of the line passing through the points
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q7.2

Question 8.
Without using the distance formula, show that the point A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q8.2
AB and CA are perpendicular to each other
Hence, ΔABC is a right-angled triangle.

Question 9.
Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q9.2
Slopes of AB and DC are equal
AB || DC Similarly slope of BC and slope of DA are equal.
BC || DA
Hence ABCD is a parallelogram.

Question 10.
(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the – quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.5
QR || PS.
Hence PQRS is a parallelogram.

Question 11.
Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.
Solution:
The given points are P (a, b + c), Q (b, c + a) and R (c, a + b)
We know that, these points P, Q, R are collinear if Slope of PQ = Slope of QR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q11.1
Slope of PQ = Slope of QR
P, Q and R are collinear.

Question 12.
Find x, if the slope of the line joining (x, 2) and (8, -11) is \(\frac { -3 }{ 4 }\).
Solution:
Slope of line joining (x, 2) and (8, -11) is
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q12.1

Question 13.
The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q13.1
Solution:
ΔABC is an equilateral
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q13.2
Each angle is equal to 60°
Side AB is parallel to x-axis
Slope of AB = slope of x-axis = 0.
Slope of AC = tan A = tan 60° = √3
Slope of CB = tan B = tan 120° = tan (180°- 60°) = – tan 60° = -√3
Slopes of AB, BC and CA are 0, -√3, √3

Question 14.
The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find :
(i) the slope of the diagonal AC
(ii) the slope of the diagonal BD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q14.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q14.2
ABCD is a square in which AB || DC || x-axis.
AD || BC || y-axis
Slope of AB and DC = 0
and slope of AD and BC = not defined (tan90° is not defined)
AC and BD are the diagonals of square ABCD.
Now slope of AC = tan 45° = 1
and slope of BD = tan 135° = tan (180° – 45°) = – tan 45° = -1

Question 15.
A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find :
(i) the slope of the altitude of AB
(ii) the slope of the median AD and
(iii) the slope of the line parallel to AC.
Solution:
Vertices of ΔABC are A (5, 4), B (-3, -2), and C (1, -8)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q15.3

Question 16.
The slope of the side BC of a rectangle ABCD is \(\frac { 2 }{ 3 }\). Find
(i) The slope of the side AB,
(ii) the slope of the side AD.
Solution:
ABCD is a rectangle in which
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q16.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q16.2

Question 17.
Find the slope and the inclination of the line AB if
(i) A = (-3, -2) and B = (1, 2)
(ii) A = (0, -√3) and B = (3, 0)
(iii) A = (-1, 2√3) and B = (-2, √3)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q17.2

Question 18.
The points (-3, 2), (2, -1) and (a, 4) are collinear, Find ‘a’.
Solution:
Points are collinear.
Slope of (-3, 2) and (2, -1) = Slope of (2, -1) and (a, 4)
Now, Slope of (-3, 2) and (2, -1) will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q18.1

Question 19.
The points (k, 3), (2, -4) and (-k + 1, -2) are collinear. Find k.
Solution:
Points (k, 3), (2, -4) and (-k + 1, -2) are collinear
Slope of (k, 3) and (2, -4) = slope of (2, -4) and (-k + 1, -2)
Now, slope of (k, 3) and (2, -4)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q19.1

Question 20.
Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.
Which segment appears to have the steeper slope, AB or AC ?
Justify your conclusion by calculating the slopes of AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q20.2

Question 21.
Find the value(s) of k so that PQ will be parallel to RS. Given :
(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)
(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)
(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q21.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q21.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A.

Other Exercises

Question 1.
Find, which of the following points lie on the line x – 2y + 5 = 0 :
(i) (1, 3)
(ii) (0, 5)
(iii) (-5, 0)
(iv) (5, 5)
(v) (2, -1.5)
(vi) (-2, -1.5)
Solution:
Equation of given line x – 2y + 5 = 0
(i) Substituting x = 1, y = 3, in the given equation.
1 – 2 x 3 + 5 = 0 ⇒ 1 – 6 + 5 = 0 ⇒ 0 = 0, which is true.
(1, 3) satisfies the equation.
(ii) Substituting x = 0 , y = 5 in the given equation
0 – 2 x 5 + 5 = 0 ⇒ 0 – 10 + 5 = 0 ⇒ -5 = 0, which is not true.
( 0, 5) does not satisfy the equation.
(iii) Substituting x = – 5, y = 0 in the given equation
-5 – 2 x 0 + 5 = 0 ⇒ -5 – 0 + 5 = 0 ⇒ 0 = 0 which is true.
(-5, 0) satisfies the equation.
(iv) Substituting x = 5, y = 5 in the given equation.
– 5 – 2 x 5 + 5 = 0 ⇒ -5 – 10 + 5 = 0 ⇒ 0 = 0 which is true.
(5, 5) satisfies the equation.
(v) Substituting x = 2, y = -1.5 in the given equation.
2 – 2 x (- 1.5) + 5 = 0 ⇒ 2 + 3 + 5 = 0 ⇒ 10 = 0. which is not true.
(2, -1.5) does not satisfy the equation.
(vi) Substituting x = -2, y = -1.5 in the given equation
– 2 – 2 x (-1.5) + 5 = 0 ⇒ – 2 + 3 + 5 = 0 ⇒ 6 = 0, which is not true.
(-2, -1.5) does not satisfies the equation.

Question 2.
State, true or false :
(i) the line \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0 passes through the point (2, 3).
(ii) the line \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0 passes through the point (4, -6).
(iii) the point (8, 7) lies on the line y – 7 = 0
(iv) the point (-3, 0) lies on the line x + 3 = 0
(v) if the point (2, a) lies on the line 2x – y = 3, then a = 5.
Solution:
(i) Equation of the line is \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0
and co-ordinates of point are (2, 3)
If the point is on the line, then it will satisfy the equation.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q2.1
(2, 3) is not on the line
(ii) Equation of the line is \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0
and co-ordinates of point are (4, -6)
If the point is on the line, then it will satisfy the equation
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q2.2
Hence, point (4, -6) is on the line.
(iii) Equation of line is y – 7 = 0 and the co-ordinates of point are (8, 7)
If the point is on the line, then it will satisfy the equation
L.H.S. = y – 7 = 7 – 7 = 0 = R.H.S.
Hence, point (8, 7) is on the line.
(iv) Equation of the line is x + 3 = 0 and co-ordinates of point are (-3, 0)
If the point is on the line, then it will satisfy the equation.
L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S.
Hence, the point (-3, 0) is on the line.
(v) Equation of the line is 2x – y = 3
and co-ordinates of the point are (2, a)
If the point is on the line, then it will satisfy the equation.
L.H.S. = 2x – y = 2 x 2 – a = 4 – a
R.H.S. = 3
4 – a = 3 ⇒ 4 + 3 = a ⇒ a = 7
But a = 5 given, therefore it is not on the line.
(i) False (ii) True (iii) True (iv) True (v) False.

Question 3.
The line given by the equation 2x – \(\frac { y }{ 3 }\) = 7 passes through the point (k, 6); calculate the value of k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q3.1

Question 4.
For what value of k will the point (3, -k) lie on the line 9x + 4y = 3 ?
Solution:
Point (3, -k) satisfies the equation 9x + 4y = 3
Substituting x = 3 , y = -k, we get :
9 x 3 + 4 (- k), = 3
⇒ 27 – 4k = 3
⇒ – 4k = 3 – 27
⇒ – 4k = – 24
⇒ k = 6

Question 5.
The line \(\frac { 3x }{ 5 }\) – \(\frac { 2y }{ 3 }\) + 1 = 0, contains the point (m, 2m – 1); calculate the value of m.
Solution:
Point (m, 2m -1) satisfies the equation
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q5.1

Question 6.
Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2) ?
Solution:
Line 3x – 5y = 6 bisect the join of points (5, -2) and (-1, 2)
The mid-point of (5, -2) and (-1, 2) satisfies the equation.
Now, mid-point of (5, -2) and (-1, 2)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q6.1
Now, substituting x = 2, y = 0, in the given equation
3 x 2 – 5 x 0 = 6 ⇒ 6 – 0 = 6 ⇒ 6 = 6 which is true. .
Given line bisect the join of points (5, -2) and (-1, 2)

Question 7.
(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of k.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and ( 0, k). Find the value of k.
Solution:
(i) line y = 3x – 2 bisects the join of (a, 3) and (2, -5)
Mid-point join of there points satisfies it.
Now, mid-point of (a, 3) and (2, -5) is
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q7.3

Question 8.
(i) The point (-3, -2) lies on the line ax + 3y + 6 = 0, calculate the value of ‘a’
(ii) The line y = mx + 8 contains the point (- 4, 4), calculate the value of ‘m’
Solution:
(i) Point (-3, 2) lies on the line ax + 3y + 6 = 0,
Then x = – 3, y = 2 satisfies it
a (-3) + 3(2) + 6 = 0
⇒ -3a + 6 + 6 = 0
⇒ -3a + 12 = 0
⇒ -3a = – 12
⇒ a = 4
(ii) line y = mx + 8 contains the point (-4, 4)
x = – 4, y = 4 satisfies it
4 = m (-4) + 8
⇒ 4 = -4m + 8
⇒ 4m = 8 – 4 = 4
⇒ m = 1

Question 9.
The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x – 5y + 15 = 0 ?
Solution:
P divides the line joining of the points (2, 1) and (-3, 6) in the ratio of 2 : 3,
co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q9.1
Now, substituting x = 0, y = 3 in the equation
x – 5y + 15 = 0
⇒ 0 – 5 x 3 + 15 = 0
⇒ 0 – 15 + 15 = 0
⇒ 0 = 0 which is true.
Point (0, 3) lies on the line.

Question 10.
The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio of 1 : 2. Does the line x – 2y = 0 contain Q ?
Solution:
Point Q, divides the line segment joining the points (5, -4) and (2, 2) in the rates of 1 : 2
co-ordinates of Q will be,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A Q10.1
Now, substituting x = 4, y = – 2 in the equation
x – 2y = 0, we get
4 – 2 x (-2) = 0
⇒ 4 + 4 = 0
⇒ 8 = 0 which is not true.
Point Q does not lie on the line x – 2y = 0

Question 11.
Find the point of intersection of the lines : 4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1) x – 2y = 4; find the value of k.
Solution:
4x + 3y = 1 …..(i)
3x – y = -9 …..(ii).
Multiplying (i) by 1 and (ii) 3
4x + 3y = 1
9x – 3y = -27
Adding, we get-:
1 3x = – 26 ⇒ x = -2
from (ii),
3x – y = – 9
3(-2) – y = -9
⇒ – 6 – y = -9
⇒ -y = -9 + 6 = -3
⇒ y = 3
The point of intersection is (-2, 3)
The line (2k – 1) x – 2y = 4 passes through that point also
It is satisfy it.
(2k – 1) (-2) – 2(3) = 4
⇒ -4k + 2 – 6 = 4
⇒ -4k – 4 = 4
⇒ -4k = 4 + 4 = 8
⇒ k = -2
Hence point of intersection is (-2, 3) and value of k = -2

Question 12.
Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Solution:
2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent
They will pass through the same point
Now 2x + 5y = 1 …..(i)
x – 3y = 6 ……(ii)
Multiply (i) by 3 and (ii) by 5, we get :
-6x + 15y = 3
5x – 15y = 30
Adding we get :
11x = 33 ⇒ x = 3
from (ii),
x – 3y = 6
⇒ 3 – 3y = 6
⇒ -3y = 6 – 3 = 3
⇒ y = -1
Point of intersection of first two lines is (3, -1)
Substituting the values in third line x + 5y + 2 = 0
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 3 – 5 + 2 = 5 – 5 = 0 = R.H.S.
Hence the given three lines are concurrent.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C.

Other Exercises

Question 1.
Given a triangle ABC in which A = (4, -4), B (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.
Solution:
B (0, 5), C (5, 10) and BP : PC = 3 : 2 Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q1.2

Question 2.
A (20, 0) and B (10, – 20) are two fixed points, find the co-ordinates of the point P in AB such that 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
Solution:
(i) A (20, 0), B (10, – 20)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q2.3

Question 3.
A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5. Show that: PQ = \(\frac { 3 }{ 8 }\) BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q3.3

Question 4.
Find the co-ordinates of points of trisection of the line segment joining the points (6, -9) and the origin.
Solution:
Points are A (6, -9) and O (0,0) let P and Q are points, which trisect AO
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q4.2

Question 5.
A line segment joining A (-1, \(\frac { 5 }{ 3 }\)) and B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects the y-axis.
(i) Calculate the value of ‘a’.
(ii) Calculate the co-ordinates of ‘P’. (1994)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q5.2

Question 6.
In what ratio is the line joining A (0, 3) and B (4, -1), divided by the x-axis ? Write the co-ordinates of the point where AB intersects the x-axis. [1993]
Solution:
Let the ratio be m1 : m2 when the x-axis intersects the line AB at P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q6.3

Question 7.
The mid point of the segment AB, as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B. (1996)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q7.1
Solution:
Let co-ordinates of A (x, 0) and B (0, y) and C (4, -3) the mid point of AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q7.3

Question 8.
AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7). Find
(i) the length of radius AC
(ii) the coordinates of B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q8.2

Question 9.
Find the co-ordinates of the centroid of a triangle ABC whose vertices are A (- 1, 3), B (1, – 1) and C (5, 1)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q9.1

Question 10.
The mid-point of the line segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
Solution:
Let A and B are two points and P is its mid point then A is (4a, 2b -3), B(-4, 2b) and P is (2, -2a)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q10.1

Question 11.
The mid point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
Solution:
The midpoint of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q11.1

Question 12.
(i) Write down the co-ordinates of the point P that divides the line joining A (-4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB ? [ICSE 1995]
Solution:
Point P, divides a line segment giving the points A (-4, 1) and B (17, 10) is the ratio 1 : 2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q12.3

Question 13.
Prove that the points A(-5, 4); B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D. So that ABCD is a square. [1992]
Solution:
In ABC, the co-ordinates of A, B and C are (-5, 4), B(-1, -2) and C (5, 2) respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q13.1
ABC is also a right-angled triangle.
Hence ABC is an isosceles right angled triangle,
Let D be the fourth vertex of square ABCD and co-ordinates of D be (x,y)
Since the diagonals of a square bisect each other and let O be the point of intersection of AC and BD.
O is mid-point of AC as well as BD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q13.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q13.4

Question 14.
M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1). Find the co-ordinates of point M. Further, if R (2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q.
Solution:
Two points are given A (-3, 7) and B (9, -1)
M is the mid-point of line joining AB.
Co-ordinates of M wll be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q14.1

Question 15.
Calculate the ratio in which the line joining A (-4, 2) and B (3, 6) is divided by point P (x, 3). Also find
(i) x
(ii) Length of AP. (2014)
Solution:
Let ratio = k : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q15.3

Question 16.
Find the ratio in which the line 2x + y = 4 divides the line segment joining the points P (2, -2) and Q (3, 7).
Solution:
Let the given line 2x + y = 4 divides the line segment joining the points P (2, -2) and Q (3,7) in the ratio k : 1 at a point (x, y) on it.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q16.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q16.2

Question 17.
If the abscissa of a point P is 2. Find the ratio in which this point divides the line segment joining the points (-4, 3) and (6, 3). Also, find the co-ordinate of point P.
Solution:
Abscissa of a point P is 2
Let co-ordinates of point P be (2, y)
Let point P (2, y) divides the line segment joining the points (-4, 3) and (6, 3) in the ratio k : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q17.1

Question 18.
The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0, find the value of k, Also, find the co-ordinates of point Q.
Solution:
A line joining the points (2, 1) and (5, -8) is trisector at P and Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q18.2

Question 19.
M is the mid-point of the line segment joining the points A (0, 4) and B (6, 0). M also divides the line segment OP in the ratio 1 : 3. Find:
(i) co-ordinates of M
(ii) co-ordinates of P
(iii) length of BP
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q19.1
Solution:
M is mid point of the line segment joining the points A (0, 4) and B (6, 0)
M divides the line segment OP in the ratio 1 : 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q19.3

Question 20.
Find the image of the point A (5, -3) under reflection in the point P (-1, 3).
Solution:
Image of the point A (5, -3) under reflection in the point P (-1, 3)
Let B (x, y) be the point of reflection of A (5, -3) under P(-1, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q20.2

Question 21.
A (-4, 2), B (0, 2) and C (-2, -4) are vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB respectively. Show that the centroid of PQR is the same as the centroid of ABC.
Solution:
A (-4, 2), B (0, 2) and C (-2, -4) are the vertices of ABC.
P, Q and R are the mid-points of the sides BC, CA and AB respectively.
G is the centroid of medians AP, BQ and CR.
Co-ordinates of G are
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q21.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C Q21.2

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