## 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

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)

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:

Solution:

Question 4.
For triangle ABC, Show that:

Solution:

Question 5.
Evaluate:

Solution:

Question 6.

Solution:

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°

(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:

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:

Question 9.
Prove that:

Solution:

Question 10.
Evaluate:

Solution:

Question 11.
Without using trigonometric tables, evaluate sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°.
Solution:

Question 12.
Without using trigonometrical tables, evaluate: cosec2 57° – tan2 33° + cos 44° cosec 46° – $$\sqrt{2}$$ cos45°- tan2 60°
Solution:

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## 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:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

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
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
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:

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
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.

Solution:

P.Q.

Solution:

P.Q.
Evaluate:

Solution:

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

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## 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.

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)

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.

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.

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)

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.

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.

To prove:
(i) CB : BA = CP : PA
(ii) AB x BC = BP x CA
Proof:
(i) In ΔABC,
BP is the bisector of ∠ABC

Question 6.
In ΔABC; BM ⊥ AC and CN ⊥ AB; show that:

Solution:
In ΔABC,
BM ⊥ AC and CN ⊥ AB
To prove:

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.

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
(c) ΔAME ~ ΔANC
(ii) ΔAME ~ ΔANC

Question 8.
In the given figure, AD = AE and AD² = BD x EC
Prove that: triangles ABD and CAE are similar.

Solution:
In the given figure,
To prove: ΔABD ~ ΔCAE
∠ADE = ∠AED (Angles opposite to equal sides)

Question 9.
In the given figure, AB // DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.

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)

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

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.

Solution:

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:

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°)

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)

Question 16.
Given : ABCD is a rhombus, DPR and CBR are straight lines.

Prove that: DP x CR = DC x PR.
Solution:
Proof: In ΔAPD and ΔPRC
∠DPA = ∠CPR (Vertically opposite angles)
ΔAPD ~ ΔPRC (AA Postulate)

Question 17.
Given: FB = FD, AE ⊥ FD and FC ⊥ AD.
Prove : $$\frac { FB }{ AD }$$ = $$\frac { BC }{ ED }$$
Solution:

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)

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)

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.

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)

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

∠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
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°

Solution:
Given: In ΔABC, AD x BC and AD² = BD x DC
To Prove: ∠BAC = 90°
Proof:

Question 23.
In the given figure AB || EF || DC; AB = 67.5 cm. DC = 40.5 cm and AE = 52.5 cm.

(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

Question 24.
In the given figure, QR is parallel to AB and DR is parallel to QB.

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)

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:

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)

Question 26.
In the figure given below P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.

(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.

Question 27.
In the right angled triangle QPR, PM is an altitude.

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)

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.

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.

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

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## 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.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

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:

Question 9.
cosec A (1 + cos A) (cosec A – cot A) = 1
Solution:

Question 10.
sec2 A + cosec2A = sec2 A cosec2 A
Solution:

Question 11.

Solution:

Question 12.
tan2A – sin2A = tan2 A. sin2 A
Solution:

Question 13.
cot2 A – cos2 A = cos2 A. cot2 A
Solution:

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:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

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:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Question 34.

Solution:

Question 35.

Solution:

Question 36.

Solution:

Question 37.

Solution:

Question 38.
(1 +cot A-cosec A) (1 + tan A + sec A) = 2
Solution:

Question 39.

Solution:

Question 40.

Solution:

Question 41.

Solution:

Question 42.

Solution:

Question 43.

Solution:

Question 44.

Solution:

Question 45.

Solution:

Question 46.

Solution:

Question 47.

Solution:

Question 48.

Solution:

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## 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.

Solution:

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.

Solution:

Question 3.
Calculate BC.

Solution:

Question 4.
Calculate AB .

Solution:

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°

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:

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 :

Solution:

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:

(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:

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:

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]

Solution:
Let CD be the height of the tower and height of window A from the ground = 10m
In right ∆AEC,

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:

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:

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,

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

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

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°

Question 16.

Solution:

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:

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 :

(i) the horizontal distance between AB and CD.
(it) the height of the lamp post.
Solution:

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:

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)

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°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C are helpful to complete your math homework.

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## 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:

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

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 + $$\frac { 4 }{ 3 }$$

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

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)

Solution:

(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)

⇒ 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)

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)

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:

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.

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:

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.

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)

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

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

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

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)

(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)

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)

⇒ 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.

⇒ 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

Slopes of PQ and BC are same.
These are parallel to each other.

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

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.

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)

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)

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

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)

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)

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.

Solution:
As P (2, -3) is mid-point of AB.
Let coordinates of B be (0, y) and coordinates of A be (x, 0)

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

⇒ 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:

(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

⇒ 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

Question 31.
Find:

(i) equation of AB
(ii) equation of CD
Solution:
Co-ordinates of A and B are (-5, 4) and (3, 3) respectively.

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

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

(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:

(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.

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
∠P = ∠P (common)
∠D = ∠C (each 90°)
PB = 2PA ⇒ PA = $$\frac { 1 }{ 2 }$$ PB

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°

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

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

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.

(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

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 –

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.

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,

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

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## 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

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:

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:

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

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

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

Question 10.
If the lines y = 3x + 7 and 2y +px = 3 are perpendicular to each other, find the value of p. (2006)
Solution:

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)

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

⇒ 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

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)

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.

Solution:
(i) From the figure, the see that co-ordinates of A are (2, 3), B are (-1, 2) and C are (3, 0)

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:

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:

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

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:

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

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)

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
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:

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]

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

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

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

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

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)

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

Question 6.
The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :
(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.

⇒ 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 :

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.

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:
D is mid point of BC

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.

Solution:

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
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)

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:

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:

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)

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

Question 16.
Find the equation of the line whose slope is $$\frac { -5 }{ 6 }$$ and x-intercept is 6.
Solution:

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)

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)

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.

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.

(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)

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

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

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

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

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:

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:

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)

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

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:

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:

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:

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

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

Question 13.
The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.

Solution:
ΔABC is an equilateral

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.

Solution:

ABCD is a square in which AB || DC || x-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)

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

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:

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

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)

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:

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:

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

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## 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.

(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

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:

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

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)

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

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

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,

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
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
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

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

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)

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:

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

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:

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.

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)

Solution:
Let co-ordinates of A (x, 0) and B (0, y) and C (4, -3) the mid point of AB.

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:

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:

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)

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)

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.

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.

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.

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

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

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.

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

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.

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

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

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)

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

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