Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A.

Other Exercises

Question 1.
State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 x 2 and 2 x 3 respectively; then their sum A + B is possible,
(ii) The matrices A 2 x3, and B 2 x 3, are conformable for subtraction,
(iii) Transpose of a 2 x 1 matrix is a 2 x 1 matrix,
(iv) Transpose of a square matrix is a square matrix,
(v) A column matrix has many columns and only one row.
Solution:
(i) False, because the orders of both the matrices are not same,
(ii) True,
(iii) False, because transpose of a 2 x 1 matrix is a 1 x 2 matrix,
(iv) True,
(v) False, because it has only one column and may have many rows.

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q2.1
Solution:
Comparing the elements in order, we get that :
x = 3
y + 2 = 1 ⇒ y = 1 – 2 = -1
z – 1 = 2 ⇒ z = 2 + 1 = 3
Hence x = 3, y = -1 , z = 3

Question 3.
Solve for a, b and c; if :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q3.1
Solution:
(i) Comparing the elements in order, we get
-4 = b + 4 ⇒ – b = 4 + 4 = 8 ⇒ b = – 8
a + 5 = 2 ⇒ a = 2 – 5= -3
2 = c – 1 ⇒ c = 2 + 1 = 3
Hence a = -3, b = -8, c = 3
(ii) Comparing the elements in order, we get
a = 3
a – b = – 1 ⇒ 3 – b = -1 ⇒ -b = -1 – 3 ⇒ -b = -4 ⇒ b = 4
b + c = 2 ⇒ 4 + c = 2 ⇒ c = 2 – 4 = -2
Hence a = 3, b = 4, c = – 2

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q4.2

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q5.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q5.2

Question 6.
Wherever possible, write each of the following as a single matrix.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q6.2
This is not possible because both the matrices are not of the same order.

Question 7.
Find, x and y from the following equations:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q7.1
(ii) [-8 x] + [y -2] = [-3 2]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q7.2
Comparing, the elements of two equal martrices:
3 – x = 7 ⇒ x = -7 + 3 = -4
and y + 2 = 2 ⇒ y = 2 – 2 = 0
Hence x = – 4, y = 0
(ii) [-8 x] + [y -2] = [-3 2]
⇒ [-8 + y x – 2] = [-3 2]
Comparing the elements, we get :
-8 + y = -3 ⇒ y = -3 + 8 = 5
x – 2 = 2 ⇒ x = 2 + 2 = 4
Hence x = 4, y = 5

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q8.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q8.3

Question 9.
Write the additive inverse of matrices A, B and C
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q9.2

Question 10.
Given A = [ 2 -3], B = [0 2] and C = [-1 4]; find the matrix X in each of the following :
(i) X + B = C – A
(ii) A – X = B + C
Solution:
A = [2 -3], B = [0 2], C = [-1 4]
(i) X + B = C – A
X + [0 2] = [-1 4] – [2 -3]
⇒ X + [0 2] = [- 1 – 2 4 – (-3)]
X + [0 2] = [-3 4 + 3] = [-3 7]
X = [-3 7] – [0 2] = [-3 – 0 7 – 2] = [-3 5]
(ii) A – X = B + C
⇒ -X = B + C – A
⇒ X = A – B – C = [2 -3] – [0 2] – [-1 4] = [2 – 0 – (-1) -3 – 2 – 4] = [ 2 + 1 – 9] = [3 – 9]

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q11.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q11.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A Q11.4

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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 15 Similarity Ex 15C

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

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 15C.

Other Exercises

Question 1.
(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.
(ii) Areas of two similar triangles are 98 sq.cm and 128 sq.cm. Find the ratio between the lengths of their corresponding sides.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q1.2

Question 2.
A line PQ is drawn parallel to the base BC, of ΔABC which meets sides AB and AC at points P and Q respectively. If AP = \(\frac { 1 }{ 3 }\) PB; find the value of:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q2.3

Question 3.
The perimeters of two similar triangles are 30 cm and 24 cm. If one side of first triangle is 12 cm, determine the corresponding side of the second triangle.
Solution:
Let we are given ΔABC and ΔPQR are similar.
Perimeter of ΔABC = 30 cm.
and perimeter of ΔPQR = 24 cm.
and side BC = 12 cm.
Now we have to find the length of QR, the corresponding side of ΔPQR
ΔABC ~ ΔPQR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q3.1

Question 4.
In the given figure AX : XB = 3 : 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q4.1
Find :
(i) the length of BC, if length of XY is 18 cm.
(ii) ratio between the areas of trapezium XBCY and triangle ABC.
Solution:
We are given in the ΔABC, AX : XB = 3 : 5
XY || BC.
Let AX = 3x and XB = 5x
AB = 3x + 5x = 8x.
Now in ΔAXY and ΔABC,
∠AXY = ∠ABC (corresponding angles)
∠A = ∠A (common)
ΔAXY ~ ΔABC (AA Postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q4.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q4.3

Question 5.
ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
Given- In ΔABC, PQ || BC in such away that area APQ = area PQCB
To Find- The ratio of BP : AB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q5.1
Solution:
In ΔABC, PQ || BC.
Z APQ = Z ABC (corresponding angles)
Now in ΔAPQ and ΔABC,
∠APQ = ∠ABC (proved)
∠A = ∠A (common)
ΔAPQ ~ ΔABC (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q5.3

Question 6.
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q6.1

Solution:
In ΔPQR, LM || QR in such away that PM : MR = 3 : 4
(i) In ΔPQR, LM || QR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q6.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q6.4

Question 7.
The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel: PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q7.1
Calculate-
(i) the length of AP
(ii) the ratio of the areas of triangle APQ and triangle ABC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q7.3

Question 8.
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm². Calculate
(i) area of triangle CDP
(ii) area of parallelogram ABCD [1996]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q8.1
Solution:
(i) Join QC.
In ΔBPQ and ΔCPD,
∠DPC = ∠BPQ (vertically opposite angles.)
∠PDC = ∠BQP (Alternate angles.)
ΔBPQ ~ ΔCDP (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q8.2
⇒ area ΔCDP = 4 (area ΔBPQ)
⇒ 2 (2 area ΔBPQ) = 2 x 20 = 40 cm² (2 area A BPQ = 20 cm²)
(ii) Area || gm ABCD = area ΔCPD + area ΔADQ – area ΔBPQ
= 40 + 9 (area BPQ) – area BPQ [(AD = CB = 3 BP)]
= 40 + 8 (area ΔBPQ)
= 40 + 8 (10) cm²
= 40 + 80
= 120 cm²

Question 9.
In the given figure. BC is parallel to DE. Area of triangle ABC = 25 cm². Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC. Also. Find the area of triangle BCD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q9.1
Solution:
In ΔADE, BC || DE
Area of ΔABC = 25 cm²
and area of trapezium BCED = 24 cm²
Area of ΔADE = 25 + 24 = 49 cm²
DE = 14 cm,
Let BC = x cm.
Now in ΔABC and ΔADE,
∠ABC = ∠ADE (corresponding angles)
∠A = ∠A (common)
ΔABC ~ ΔADE (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q9.2

Question 10.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q10.1
Find-
(i) ΔAPB : ΔCPB
(ii) ΔDPC : ΔAPB
(iii) ΔADP : ΔAPB
(iv) ΔAPB : ΔADB
Solution:
AP : CP = 3 : 5 ⇒ \(\frac { AP }{ CP }\) = \(\frac { 3 }{ 5 }\)
(i) Now in ΔAPB and ΔCPB,
These triangles have same vertex and their bases are in the same straight line
area ΔAPB : area ΔCPB = AP : PC = 3 : 5 or ΔAPB : ΔCPB = 3 : 5
(ii) In ΔAPB and ΔDPC,
∠APB = ∠DPC (vertically opposite angles)
∠PAB = ∠PCD (alternate angles)
ΔAPB ~ ΔDPC (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q10.2
⇒ area ΔDPC : area ΔAPB = 25 : 9 or ΔDPC : ΔAPB = 25 : 9
(iii) In ΔADP and ΔAPB,
There have the same vertex and their bases arc in the same straight line.
area ΔADP : area ΔAPB = DP : PB
But PC : AP = 5 : 3
ΔADP : ΔAPB = 5 : 3
(iv) Similarly area ΔAPB : area ΔADB = PB : DB = 3 : (3 + 5) = 3 : 8

Question 11.
In the given figure, ARC is a triangle. DE is parallel to BC and \(\frac { AD }{ DB }\) = \(\frac { 3 }{ 2 }\).
(i) Determine the ratios \(\frac { AD }{ AB }\) , \(\frac { DE }{ BC }\)
(ii) Prove that ΔDEF is similar to ΔCBF.
Hence, find \(\frac { EF }{ FB }\)
(iii) What is the ratio of the areas of ΔDEF and ΔBFC?
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q11.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q11.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q11.4

Question 12.
In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ΔABC and ΔDEC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q12.1
Solution:
In the figure DE = 7.8 cm, AB = 10.4 cm
∠ACD = ∠BCE (given)
Adding ∠DCB both sides,
∠ACD + ∠DCB = ∠DCB + ∠BCE
∠ACB = ∠DCE
Now in ΔABC and ΔDCE
∠B = ∠E (given)
∠ACB = ∠DCE (proved)
ΔABC ~ ΔDCE (AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q12.2

Question 13.
Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q13.1
Solution:
In the figure, ΔABC is an isosceles triangle in which
AB = AC = 13 cm, BC = 10 cm,
AD ⊥ BC, CE = 8 cm and EF ⊥ AB
(i) Now in ΔADC and ΔFEB
∠C = ∠B (AB = AC)
∠ADC = ∠EFB (each = 90°)
ΔADC ~ ΔFEB (AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15C Q13.3

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

Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21D

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 21D.

Other Exercises

Question 1.
Use tables to find sine of:
(i) 21°
(ii) 34°42′
(iii) 47° 32′                    
(iv) 62°57′
(v) 10°20′ + 20° 45′
Solution:
From tables of sine of angles, we find that:
(i) sin 21°= 0.3584,
(ii) sin 34°42’= .5693
(iii) sin 47° 32′ = 0.7377
(iv) sin 62° 57′ = 0.8906
(v) sin 10° 20′ + 20°45′ = sin 31°5′
= 0.5162

Question 2.
Use tables to find cosine of:
(i) 2°4′
(ii) 8°12′
(iii) 26°32’                     
(iv) 65°41′
(v) 9°23′ +15°54′
Solution:
From tables of cosine of angle, we find that:
(i) cos 2°4′ = 0.9993
(ii) cos 8° 12’ = 0.9898
(iii) cos 26°32′ = 0.8946
(iv) cos 65°41′ = 0.4118
(v) cos 9°23′ + 15°54′ = cos 25° 17′
= 0.9042

Question 3.
Use trigonometrical tables to find tangent of:
(i) 37°
(ii) 42°18′
(iii) 17°27′
Solution:
From the tables of tangents, we find that
(i) tan 35° = 0.7536
(ii) tan 42°18’= 0.9099
(iii) tan 17°27’= 0.3144

Question 4.
Use tables to find the acute angle θ, if the value of sin θ
(i) 4848
(ii) 0.3827
(iii) 0.6525
Solution:
From the tables of series, we find that of :
(i) sinθ = 0.4848, then θ = 29°
(ii) sinθ = 0.3827, then θ = 20° 30′
(iii) sin θ = 0.6525, then θ = 40° 42’ + 2′ = 40°44′

Question 5.
Use tables to find the acute angle θ, if the value of cos θ is :
(i) 0.9848
(ii) 0.9574
(iii) 0.6885
Solution:
From the tables of cosines, we find that if :
(i) cos θ = 0.9848, then θ = 10°
(ii) cos θ = 0.9574, then θ = 16°48′- 1’=16°47’
(iii) cos θ = 0.6885, then θ = 46° 30′ or 46°30′
= 46° 29’

Question 6.
Use tables to find the acute angle θ, if the value of tan θ is :
(i) 2419
(ii) 0.4741
(iii) 0.7391                     
(iv) 1.06
Solution:
From the table of tangents, we find that if:
(i) tan θ = 0.2419, then θ=13° 36’
(ii) tan θ = 0.4741, then θ = 25° 18’ + 4’ = 25°22′
(iii) tan θ = 0.7391, then θ= 36°24’+ 4′ = 36°28′
(iv) tan θ = 1.06, then θ = 46°36′ + 4′ = 46°40′

Question 7.
If sin θ=0.857; find:
(i) θ                              
(ii) tan θ
Solution:
From the tables of T. Ratio’s we find this :
(i) If sin θ = 0.857, then θ = 58°54′ + 4.5′ = 58° 58′ or 58°59’
(ii) tan 58°58’= 1.6577 +43 = 1.662 or tan 58° 59′ = 1.6577 + 53 = 1.663

Question 8.
If θ is the acute angle and cos θ = 0.7258; find:
(i) θ
(ii) 2 tan θ – sin θ
Solution:
From the tables of T-ratio’s, we find that:
(i) If cos θ = 0.7258, then θ= 43° 30′ -2′ = 43°28’
(ii) Now 2 tan θ – sin θ= 2 tan 43°28′ – sin 43°28′
2 tan 43°28’ = 2 x (0.9457 + 0.0022)
= 0:9479 x 2 = 1.8958
and sin 43°28′ = 0.6871 + 0.0008 = 0.6879
∴ 2 tan 43°28′ – sin 43° 28′ = 1.8958 – 0.6879 = 1.2079

Question 9.
Let θ be an acute angle and tan θ = 0.9490 find:
(i) θ
(ii) cos θ
(iii) sin θ – cos θ

Solution:
From the tables of T-raios, we find that:
(i) if tan θ = 0.9490 , then θ = 43°30′
(ii) cos θ = cos 43°30′ = 0.7254
(iii) sin θ = sin 43°50′ = 0.6884
∴ sin θ – cos θ = 0.6884 – 0.7254 = -0.0370 = -0.037

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

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

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 15B.

Other Exercises

Question 1.
In the following figure, point D divides AB in the ratio 3 : 5. Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.4

Question 2.
In the given figure, PQ // AB;
CQ = 4.8 cm, QB = 3.6 cm and AB = 6.3 cm. Find:
(i) \(\frac { CP }{ PA }\)
(ii) PQ
(iii) If AP = x, then the value of AC in terms of x.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q2.1
Solution:
In the given figure,
PQ || AB
CQ = 4.8 cm, QB = 3.6 cm, AB = 6.3 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q2.3

Question 3.
A line PQ is drawn parallel tp the side BC of ΔABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.
Solution:
In ΔABC, PQ || BC
AB = 9.0 cm, CA = 6 cm, AQ = 4.2 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q3.1

Question 4.
In ΔABC, D and E are the points on sides AB and AC respectively.
Find whether DE // BC, if:
(i) AB = 9 cm, AD = 4 cm, AE = 6 cm and EC = 7.5 cm
(ii) AB = 63 cm, EC = 11.0 cm, AD = 0.8 cm and AE = 1.6 cm.
Solution:
In ΔABC, D and E are the points on sides AB and AC respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q4.2

Question 5.
In the given figure, ΔABC ~ ΔADE. If AE : EC = 4 : 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of ‘x’.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q5.1
Solution:
In the given figure,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q5.2
ΔABC ~ ΔADE
AE : EC = 4 : 7, DE = 6.6 cm, BC = ?
Draw AL ⊥ DE and AM ⊥ BC and AL = x cm
Find AM in terms of x
ΔADE ~ ΔABC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q5.3

Question 6.
A line segment DE is drawn parallel to base BC of ΔABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm, find the length of AE.
Solution:
In ΔABC DE || BC
AB = 5 BD, EC = 3.2 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q6.2

Question 7.
In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3 cm, calculate the values of x and y.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q7.1
Solution:
(i) In ΔACB and ΔFCE, we have
∠ACB = ∠FCE (vertically opposite angles)
∠CBA = ∠CEF (alternate angles)
ΔACB ~ ΔFCE (AA Axiom of similarity)
Thus their corresponding sides are proportional.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q7.3

Question 8.
In the figure, given below, PQR is a right angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q8.1
Solution:
Given, PQ = 6 cm; PY = 4 cm;
PX : XQ = 1 : 2
Since a line drawn || to one side of triangle divide the other two sides proportionally.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q8.2

Question 9.
In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that: PE = 2PD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q9.1
Solution:
In the given figure, ABCD is a ||gm
AB || CD, AD || BC
M is mid point of BC
DM intersect AB produced at E and AC at P
To prove: PE = 2PD
Proof: In ΔDEA,
AD || BC (Opposite sides of || gm)
M is mid-point of CB B is mid-point of AE
AB = BE ⇒ AE = 2AB or 2CD
In ΔPAE and ΔPCD
∠APE = ∠CPD (Vertically opposite angles)
∠PAE = ∠PCD (Alternate angles)
ΔPAE ~ ΔPCD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q9.2

Question 10.
The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q10.1
Solution:
In the given figure, ABCD is a ||gm E is a point on AD
CE is produced to meet BA produced at point F
AE = 4 cm, AF = 8 cm, AB = 12 cm
To find the perimeter of ||gm ABCD
In ΔFBC,
AD or AE || BC (Opposite sides of ||gm)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q10.2
Perimeter of ||gm ABCD = 2 (AB + BC) = 2 (12 + 10) cm = 2 x 22 = 44 cm

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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 13 Section and Mid-Point Formula Ex 13A

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

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 13A.

Other Exercises

Question 1.
Calculate the co-ordinates-of the point P which divides the line segment joining:
(i) A (1, 3) and B (5, 9) in the ratio 1 : 2
(ii) A (-4, 6) and B (3, -5) in the ratio 3 : 2
Solution:
(i) Let co-ordinates of P be (x,y)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q1.2

Question 2.
In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis ?
Solution:
Let the point P (x, 0) divides in the ratio of m1 : m2 line joining the points A (2, -3) and B (5, 6)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q2.1

Question 3.
In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis ?
Solution:
Let the point P (0, y) divides the line joining the points A (2, -4) and (-3, 6) in the ratio of m1 : m2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q3.1

Question 4.
In what ratio does the point (1, a) divide the join of (-1, 4) and (4, -1)? Also, find the value of ‘a’.
Solution:
Let the point P (1, a) divides the line joining the points (-1, 4) and (4, -1) in the ratio of m1 : m2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q4.1

Question 5.
In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8) ? Also, find the value of ‘a’.
Solution:
Let the point P (a, 6) divides the line joining the points A (-4, 3), B (2, 8) in the ratio of m1 : m2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q5.1

Question 6.
In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q6.2

Question 7.
Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the co-ordinates of the point of intersection.
Solution:
Let, the points (0, y) be the point of intersection which divides the line joining the points A (-4, 7) and B (3, 0)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q7.1

Question 8.
Points A, B, C and D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of A, B, C and D.
Solution:
Points A, B, C and D divide the line segment joining the points (5, -10) and origin (0, 0) in five equal parts
Let co-ordinates of A be (x, y) which divides PO in the ratio of 1 : 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q8.3

Question 9.
The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that = \(\frac { PA }{ PB }\) = \(\frac { 1 }{ 5 }\), find the co-ordinates of P.
Solution:
Let the co-ordinates of P be (x, y) which divides the line joining the points A (-3,-10) and B (-2,6) in the ratio of AP : PB i.e. (5 – 1) : 1 or 4 : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q9.1

Question 10.
P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q10.1

Question 11.
Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.
Solution:
Let the point P (2, y) divides the line joining the points A (-3, -1) and B (5, 7) in the ratio of m1 : m2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q11.1

Question 12.
Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2.
Solution:
Let the point P (x, 2) divides the line joining the points A (6, 5) and B (4, -3) in the ratio of m1 : m2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q12.1

Question 13.
The point P(5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find the co-ordinates of points A and B.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q13.1
Solution:
From the figure, the line AB intersects x-axis at A and y-axis at B.
Let the co-ordinates of A (x, 0) and B (0, y) and P (5, -4) divides it in the ratio of 2 : 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q13.2

Question 14.
Find the co-ordinates of the points of trisection of the line joining the points (-3, 0) and (6, 6).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q14.2

Question 15.
Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q15.1
Let the points A (-5, 8) and B (10, -4).
Let P and Q be the two points on the axis which trisect the line joining the points A and B.
AP = PQ = QB
AP : PB = 1 : 2 and AQ : QB = 2 : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q15.3

Question 16.
Show that A (3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other point of trisection.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q16.1
Let A and B are the points of trisection of the line segment joining the points P (2, 1) and Q (5, -8), then
PA = AB = BQ.
PA : AQ = 1 : 2 and PB : BQ = 2 : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q16.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q16.3

Question 17.
If A = (-4, 3) and B = (8, -6)
(i) find the length of AB
(ii) In what ratio is the line joining A and B, divided by the x-axis ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q17.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q17.3

Question 18.
The line segment joining the points M (5, 7) and N (-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q18.1

Question 19.
A (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P and Q lie on AB and AC respectively, such that:
AP : PB = AQ : QC = 1 : 2.
(i) Calculate the co-ordinates of P and Q.
(ii) Show that PQ = \(\frac { 1 }{ 3 }\) BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q19.2

Question 20.
A (-3, 4), B ( 3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP: PC = 2 : 3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q20.2

Question 21.
ThelinesegmentjoiningA(2, 3)andB(6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the pointK. [1990, 2006]
Solution:
Let the line segment Intersect the x-axis at the point P
Co-ordinates of P are (x, 0)
Let P divide the line segment in the ratio K : 1 then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q21.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q21.2

Question 22.
The line segment joining A (4, 7) and B (-6, -2) is intercepted by the y-axis at the point K. Write down the abscissa of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Solution:
Points A (4, -7), B (-6, -2) are joined which intersects y-axis at K. abscissa of K will be 0
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q22.1
Let the coordinates of K be (0, y) and K divides AB line segment in the ratio m1 : m2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q22.2

Question 23.
The line joining P (-4, 5) and Q (3, 2), intersects they axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find:
(i) The ratio PR: RQ.
(ii) The co-ordinates of R.
(iii) The areas of the quadrilateral PMNQ. [2004]
Solution:
(i) Let divides the line joining the points P (-4, 5) and Q (3, 2) in the ratio k : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q23.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q23.2

Question 24.
In the given figure, line APB meets the x- axis at point A and y-axis at point B. P is the point (-4, 2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q24.1
Solution:
Let the co-ordinates of A be (x1, 0) (as it lies on x-axis)
and co-ordinates of B be (0, y2)
and co-ordinates of P are (-4, 2)
AP : PB = 1 : 2 i.e. m1 = 1, m2 = 2
Now, P divides AB in the ratio m1 : m2 or 1 : 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q24.2

Question 25.
Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q25.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A Q25.2

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives) Ex 23

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives) Ex 23.

Question 1.
Draw histograms for the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.4

Question 2.
Draw a cumulative frequency curve (ogive) for each of the following distributions
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q2.2

Question 3.
Draw an ogive for each of the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.2
Now plotting the points (10, 8), (20, 25), (30, 38), (40, 50) and (50, 67) on the graph and join them with free hand to obtain an ogive as shown in the graph.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.3
Now plot the points (10, 0), (20, 17), (30, 32), (40, 37), (50, 53), (60, 58) and (70, 65) on the graph and join them with free hand to get an ogive as shown in the graph.

Question 4.
Construct a frequency distribution table for the numbers given below, using the class intervals 21-30,31-40…. etc.
75, 67, 57, 50, 26, 33, 44, 58, 67,75, 78, 43, 41, 31, 21, 32, 40, 62, 54, 69, 48, 47,51,38, 39,43,61, 63, 68, 53, 56, 49, 59, 37, 40, 68, 23, 28, 36 and 47.
Use the table obtained to draw:
(i) a histogram
(ii) an ogive
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.3
(ii) Ogive :
We plot the points (30, 4), (40, 13), (50, 22), (60, 29); (70, 37) and (80, 40) on the graph and join them in free hand to obtain an ogive.

Question 5.
(a) Use the information given in the adjoining histogram to c construct a f frequency table.
(b)Use this table to construct an ogive.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.1
Solution:
From the histogram given, the required frequency table will be as given below.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.2
Plot the points (12, 9), (16, 25), (20, 47), (24, 65), (28, 77), (32, 81) on the graph and join them with free hand to get an ogive as shown.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.3

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.1
(a) From the distribution given above, construct a frequency table, (b) Use the table obtained in part (a) to draw :
(i) a histogram
(ii) an ogive.
Solution:
Difference in consecutive class marks. = 17.5 – 12.5 = 5
∴ first class interval will be : 10-15 and so on
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.2
(ii) We plot the point (15. 12), (20. 29), (25. 51), (30, 78). (35, 108), (40. 129) and (45, 145) on the graph and join them in free hand to obtain the ogive.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.3

Question 7.
Use graph paper for this question.
The table given beiow shows the monthly wages of some factory workers.
(i) Using the table calculate the cumulative frequencies of workers.
(ii) Draw the cumulative frequency curve.
Use 2 cm = Rs. 500, starting the origin at Rs. 6500 on x-axis, and 2 cm = 10 workers on the y-axis
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.3
(ii) We plot the points (7000, 10), (7500, 28), (8000, 50), (8500, 75), (9000, 92) (9500, 102) and (10000, 110) on the graph and join them in free hand to obtain an ogive.

Question 8.
The following table shows the distribution of the heights of a group of factory workers :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.1
(i) Determine the cumulative frequencies.
(ii) Draw the ‘less than’ cumulative frequency curve on a graph paper. Use 2 cm = 5cm height on one axis and 2 cm = 10 workers on the other.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.3
We plot points (155, 6), (160, 18), (165, 36),
(170, 56), (175, 69), (180, 77) and (185, 83) on the graph and join them in free hand to obtain an ogive.

Question 9.
Construct a frequency distribution table for each of the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q9.2

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B.

Other Exercises

Question 1.
Using the Factor Theorem, show that:
(i) (x – 2) is a factor of x3 – 2x2 – 9x +18. Hence, factorise the expression x3 – 2x2 – 9x + 18 completely.
(ii) (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
(iii) (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2. completely.
(iv) 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence, factorise the given expression completely.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q1.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q1.5

Question 2.
Using the Remainder Theorem, factorise each of the following completely:
(i) 3x3 + 2x2 – 19x + 6
(ii) 2x3 + x2 – 13x + 6
(iii) 3x3 + 2x2 – 23x – 30
(iv) 4x3 + 7x2 – 36x – 63
(v) x3 + x2 – 4x – 4. (2004)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q2.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q2.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q2.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q2.6

Question 3.
Using the Remainder Theorem factorise the expression 3x3 + 10x2 + x – 6. Hence, solve the equation. 3x3 + 10x2 + x – 6 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q3.3

Question 4.
Factorise the expression f(x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = o.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q4.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q4.3

Question 5.
Given that x – 2 and x +1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q5.3

Question 6.
The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q6.3

Question 7.
If x + a is a common factor of expressions f(x) = x2 + px + q and g (x) = x2 + mx + n show that a = \(\frac { n – q }{ m – p }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q7.1

Question 8.
The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q8.1

Question 9.
Find the value of ‘a’ if (x – a) is a factor of x3 – ax2 + x + 2.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q9.1

Question 10.
Find the number that must be subtracted from the polynomial 3y3 + y2 – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B Q10.1
⇒ 9 – k = 0 ⇒ k = 9
Hence 9 should be subtracted.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A.

Other Exercises

Question 1.
Find in each case, the remainder when :
(i) x4 – 3x2 + 2x + 1 is divided by x – 1.
(ii) x3 + 3x2 – 12x + 4 is divided by x – 2.
(iii) x4 + 1 is divided by x + 1.
(iv) 4x3 – 3x2 + 2x – 4 is divided by 2x + 1.
(v) 4x3 + 4x2 – 21x + 16 is divided by 2x – 3.
(vi) 2x3 + 9x2 – x – 15 is divided by 2x + 3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q1.4

Question 2.
Show that:
(i) x – 2 is a factor of 5x2 + 15x – 50.
(ii) 3x + 2 is a factor of 3x2 – x – 2.
(iii) x + 1 is a factor of x3 + 3x2 + 3x + 1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q2.2

Question 3.
Use the Remainder Theorem to find which of the following is a factor of 2x3 + 3x2 – 5x – 6.
(i) x + 1
(ii) 2x – 1
(iii) x + 2
(iv) 3x – 2
(v) 2x – 3.
Solution:
(i) f(x) = 2×3 + 3×2 – 5x – 6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q3.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q3.4

Question 4.
(i) If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2 + 2x – k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q4.2

Question 5.
Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression, x3 + ax2 + bx – 12.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q5.1
(x + 3) is a factor of f(x)
f(-3) = 0
9a – 3b – 39 – 0
⇒ 3a – b – 13 = 0
⇒ 3a – b = 13
Adding (i) and (ii) we get:
5a = 15 ⇒ a = 3
Substituting the value of a in (i)
2(3) + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6 = – 4
Hence a = 3, b = – 4

Question 6.
Find the value of k, if 2x + 1 is a factor of (3k + 2) x3 + (k – 1).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q6.2

Question 7.
Find the value of a, if x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q7.1

Question 8.
Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1) x2 + nx – 18.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q8.2

Question 9.
When x3 + 2x2 – kx + 4 is divided by x – 2, the remainder is k. Find the value of constants.
Solution:
f(x) = x3 + 2x2 – kx + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q9.1

Question 10.
Find the value of a, if the division of ax3 + 9x2 + 4x – 10 by x + 3 leaves a remainder 5.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q10.1

Question 11.
If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b. [2005]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q11.1

Question 12.
The expression 2x3 + ax2 + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q12.2

Question 13.
What number should be added to 3x3 – 5x2 + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8 ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q13.1

Question 14.
What number should be subtracted from x3 + 3x2 – 8x + 14 so that on dividing it by x – 2, the remainder is 10.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q14.1

Question 15.
The polynomials 2x3 – 7x2 + ax – 6 and x3 – 8x2 + (2a + 1) x – 16 leave the same remainder when divided by x – 2. Find the value of ‘a’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q15.1

Question 16.
If (x – 2) is a factor of the expression 7x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q16.1

Question 17.
Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leave the same remainder when divided by x + 3. (2015)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A Q17.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A 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 7 Ratio and Proportion Ex 7D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D.

Other Exercises

Question 1.
If a : b = 3 : 5, find : (10a + 3b) : (5a + 2b)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q1.1

Question 2.
If 5x + 6y : 8x + 5y = 8 : 9, find x : y
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q2.1

Question 3.
If k (3x – 4y) : (2x – 3y) = (5x – 6y) : (4x – 5y), find x : y.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q3.1

Question 4.
Find the :
(i) duplicate ratio of 2√2 : 3√5
(ii) triplicate ratio of 2a : 3b,
(iii) sub-duplicate ratio of 9x2 a4 : 25y6 b2
(iv) sub-triplicate ratio of 216 : 343
(v) reciprocal ratio of 3 : 5
(vi) ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q4.1

Question 5.
Find the value of x, if :
(i) (2x + 3) : (5x – 38) is the duplicate ratio of √5 : √6
(ii) (2x + 1) : (3x + 13) is the sub-duplicate ratio of 9 : 25.
(iii) (3x -7): (4x + 3) is the sub-triplicate ratio of 8 : 27.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q5.2

Question 6.
What quantity must be added to each term of the ratio x : y so that it may become equal to c : d ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q6.1

Question 7.
A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 84 kg?
Solution:
Ratio in reduction of weight = 7 : 5
Originally weight of the woman = 84 kg
Reduced weight = \(\frac { 84 x 5 }{ 7 }\) = 60 kg

Question 8.
If 15 (2x² – y²) = 7xy, find x : y, if x and y both are positive.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q8.2

Question 9.
Find the :
(i) fourth proportional to 2xy, x² and y².
(ii) third proportional to a² – b² and a + b.
(iii) mean proportion to (x – y) and (x3 – x²y)
Solution:
(i) Let a be the fourth proportional
Then 2xy : x² :: y² : a
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q9.1

Question 10.
Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
Solution:
Let x and y be the two numbers.
Then 14 is the mean proportional between x and y
xy = 14² => xy = 196 ….(i)
and 112 is the third proportional to x, y
y² = 112 x ….(ii)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q10.1

Question 11.
If x and y be unequal and x : y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q11.1

Question 12.
If q is the mean proportional between p and r, prove that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q12.3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q11.2

Question 13.
If a, b and c are in continued proportion, prove that: a : c = (a² + b²) : (b² + c²).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q13.1

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q14.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q14.2

Question 15.
If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that a : b = c : d.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q15.1

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q16.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q16.2

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q17.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q17.2

Question 18.
There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3 : 1. How many more girls should be added to the council so that the ratio of number of boys to the number of girls may be 9 : 5 ?
Solution:
Total number of members = 36
Ratio in boys and girls = 3 : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q18.1

Question 19.
If 7x – 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q19.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q19.3

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q20.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q20.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q20.3

Question 21.
If x, y, z arc in continued proportion, prove that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q21.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q21.2

Question 22.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q22.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q22.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q22.3

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q23.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q23.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q23.3

Question 24.
Using componendo and dividendo, find the value
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q24.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q24.2

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q25.2

Question 26.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q26.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q26.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q26.3

Question 27.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q27.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q27.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q27.3

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