OP Malhotra Class 10 Maths Solutions Chapter 11 Coordinate Geometry Ex 11(a)

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S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(a)

Question 1.
Find the mid-points of lines joining
(a) (5, 8), (9, 11)
(b) (0, 0), (8, – 5)
(c) (-7, 0), (0, 10)
(d) (-4, 3), (6, -7)
Solution:
We know that the co-ordinates of a mid-point of a line whose vertices are \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\). Therefore
(a) Mid point of the line joining (5, 8) and (9, 11) will be = \(\left(\frac{5+9}{2}, \frac{8+11}{2}\right)\)

(b) will be = Mid point of the line joining (0, 0) and (8, 5) will be = \(\left(\frac{0+8}{2}, \frac{0-5}{2}\right) \text { or }\left(4, \frac{-5}{2}\right)\)

(c) Mid point of the line joining (-7, 0) and (0, 10) will be = \(\left(\frac{-7+0}{2}, \frac{0+10}{2}\right) \text { or }\left(\frac{-7}{2}, \frac{10}{2}\right)\) or \(\left(\frac{-7}{2}, \frac{10}{2}\right)\) or \(\left(\frac{-7}{2}, 5\right)\)

(d) Mid point of the line joining (-4, 3) and (6,-7) will be = \(\left(\frac{-4+6}{2}, \frac{3-7}{2}\right) \text { or }\left(\frac{2}{2}, \frac{-4}{2}\right)\) or (1,-2)

Question 2.
Find the mid-points of the sides of a triangle whose vertices are A (1, -1), B (4, -1), C (4, 3).
Solution:
Let D, E and F be the mid-points of the sides
BC, CA and AB of the triangle ABC
∴ Co-ordinates of D, the mid-point of BC will
be \(\left(\frac{4+4}{2}, \frac{-1+3}{2}\right) \text { or }\left(\frac{8}{2}, \frac{2}{2}\right)\) or (4, 1)

Co-ordinates of E, the mid-point of CA will
be \(\left(\frac{4+1}{2}, \frac{3-1}{2}\right) \text { or }\left(\frac{5}{2}, \frac{2}{2}\right) \text { or }\left(\frac{5}{2}, 1\right)\)
and co-ordinates of F, the mid-point of AB
will be = \(\left(\frac{1+4}{2}, \frac{-1-1}{2}\right) \text { or }\left(\frac{5}{2}, \frac{-2}{2}\right)\) or \(\left(\frac{5}{2},-1\right)\)
Co-ordinates of mid-point of AB, BC and CA
are (\(\frac { 5 }{ 2 }\), – 1), (4, 1), (\(\frac { 5 }{ 2 }\), 1)

Question 3.
Find the centre of a circle if the end points of a diameter are A (-5, 7) and C (3, -11).
Solution:
Let O be the centre of the circle

∴ O will be the mid point of the diameter AC,
let co-ordinates of O be (x, y), then
x = \(\frac{x_1+x_2}{2}, y=\frac{y_1+y_2}{2}\)
∴ x = \(\frac{-5+3}{2}, y=\frac{7-11}{2} \Rightarrow x=\frac{-2}{2}, y=\frac{-4}{2}\)
⇒ x = – 1, y = – 2
∴ Co-ordinates of O are (- 1, – 2)

Question 4.
If M is the mid-point of AB, find the co-ordinates of :
(a) A if the coordinates of M and B are M (2, 8) and B (-4, 19) and
(b) B if the co-ordinates of A and M are A (-1, 2), M (-2, 4).
Solution:
M is mid-point of AB
(a) Co-ordinates of M are (2, 8) and of B are (-4, 19)
Let co-ordinates of A be (x₁, y₁)
∴ Co-ordinates of M will be
= \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
(2, 8) = \(\left(\frac{x_1-4}{2}, \frac{y_1+19}{2}\right)\)
Comparing, we get
\(\frac{x_1-4}{2}\) = 2 ⇒ x₁ – 4 = 4 ⇒ x₁ = 4 + 4 ⇒ x₁ = 8
and \(\frac{y_1+19}{2}\) = 8 ⇒ y₁ + 19 = 16 ⇒ y₁ = 16 – 19 = – 3
∴ Co-ordinates of A are (8, -3).

(b) Let co-ordinates of B be (x₂, y₂)
But co-ordinates of M are (-2, 4) and of A are (-1,2)
∵ M is the mid point of AB and let co-ordinates of
M be \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
∴ (-2, 4) = \(\left(\frac{-1+x_2}{2}, \frac{2+y_2}{2}\right)\)
Comparing, we get
\(\frac{-1+x_2}{2}\) – 2 ⇒ – 1 + x₂ = – 4
⇒ x = – 4 + 1 = – 3
Ax = – 3 and \(\frac{2+y_2}{2}\) = 4 ⇒ 2 + y₂ = 8
⇒ y₂ = 8 – 2 = 6
∴ Co-ordinates of B are, (- 3, 6)

Question 5.
Find the coordinates of the point which divides internally the join of the points
(a) (8, 9) and (-7, 4) in the ratio 2 : 3
(b) (1, -2) and (4, 7) in the ratio 1 : 2.
Solution:
If a point (x, y) divides a line segment having
its end points (x₁, y₁) and (x₂, y₂) in the ratio
of m₁ : m₂, then
x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\) and y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}\)

(a) Let co-ordinates of the point P be (x, y) which
divides the join of points A (8, 9) and B (- 7, 4) in the ratio 2 : 3, then
m₁ = 2 and m₂ = 3
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{2 \times(-7)+3 \times 8}{2+3}\)
= \(\frac{-14+24}{5}=\frac{10}{5}=2\)
y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{2 \times 4+3 \times 9}{2+3}=\frac{8+27}{5}\)
= \(\frac { 35 }{ 5 }\) = 7
∴ Co-ordinate of the required point P will be (2, 7)

(b) Let co-ordinates of the point P be (x, y) which divides the join of points A (1, -2) and B (4, 7) in the ratio 1 : 2, then
m₁ = 1, m₂ = 2

∴ Co-ordinates of the required point will be (2, 1)

Question 6.
Find the co-ordinates of the points of trisection of the line joining the points (2, 3) and (6, 5).
Solution:
Let P and Q are the points which trisect the line segment joining the points A (2, 3) and B (6, 5)

Let co-ordinates of P be (x’, y’) and of Q be be (x”, y”), then
In case of P, m₁ = 1, m₂ = 2

∴ Co-ordinates of P will be (\(\frac { 10 }{ 3 }\), \(\frac { 11 }{ 3 }\))
In case of Q, m₁ = 2, m₂ = 1

∴ Co-ordinates of Q will be (\(\frac { 14 }{ 3 }\), \(\frac { 13 }{ 3 }\))

Question 7.
In what ratio is the line joining the points
(a) (2, -3) and (5, 6) divided by the x-axis;
(b) (3, -6) and (-6, 8) divided by they-axis?
Solution:
(a) Let the ratio in which x-axis divides the join of points (2, -3) and (5, 6), be m₁ : m₂
Let the co-ordinates of point at x-axis be (x, 0)

(b) Let the ratio in which y-axis divides the join of points (3, -6) and (-6, 8) be m₁ : m₂ and let the co-ordinates of point at y-axis be (0, y)

Question 8.
Find the centroid of the triangle whose angular points are (-4, 6), (2, -2) and (2, 5) respectively.
Solution:
Let G be the centroid of the triangle whose vertices are A (-4, 6), B (2, -2) and C (2, 5)
∴ Co-ordinates of centroid will be
\(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
or \(\left(\frac{-4+2+2}{3}, \frac{6-2+5}{3}\right)\)
or \(\left(\frac{0}{3}, \frac{9}{3}\right)\) or (0, 3)

Question 9.
If (x₁, y₁) = (2, 3); x₂ = 3 and y₃ = – 2 and G is (0, 0), find y₂ and x₃.
Solution:
∵ G is the centroid of the triangle
∴ Co-ordinates of G will be
\(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
But co-ordinates of centroid are (0, 0)

Question 10.
The mid-points of the sides of a triangle are at (1, 4), (4, 8) and (5, 6). Find the coordinates of the vertices of the triangle.
Solution:
Let D, E and F are the mid-points of a ∆ABC
Let vertices of A, B and C be (x₁, y₁), (x₂, y₂)
and (x₃, y₃) and co-ordinates of D, E and F
are (1, 4), (4, 8) and (5, 6) respectively..
∴ 1 = \(\frac{x_1+x_2}{2}\) ⇒ x₁ + x₂ = 2 … (i)
4 = \(\frac{x_2+x_3}{2}\) ⇒ x₂ + x₃ = 8 … (ii)
5 = \(\frac{x_3+x_1}{2}\) ⇒ x₃ + x₁ = 10 … (iii)
Adding we get, 2 (x₁ + x₂ + x₂) = 20
⇒ x₁ + x₂ + x₃ = \(\frac { 20 }{ 2 }\) = 10 … (iv)
Subtracting (ii), (iii) and (i) from (iv) term by term, we get
x₁ = 10 – 8 = 2
x₂ = 10 – 10 = 0
x₃ = 10 – 2 = 8
Similarly
4 = \(\frac{y_1+y_2}{2}\) ⇒ y₁ + y₂ = 8 … (v)
8 = \(\frac{y_2+y_3}{2}\) ⇒ y₂ + y₃ = 16 … (vi)
6 = \(\frac{y_3+y_1}{2}\) ⇒ y₃ + y₁ = 12 … (vii)
Adding we get,
2 (y₁ + y₂ + y₃) = 8 + 16 + 12 = 36
⇒ y₁ + y₂ + y₃ = \(\frac { 36 }{ 2 }\) = 18 (viii)
Subtracting (vi), (vii) and (v) from (viii) term by term, we get
y₁ = 18 – 16 = 2
y₂ = 18 – 12 = 6
y₃ = 18 – 8 = 10
∴ Co-ordinates of A, B and C are (2, 2), (0, 6), (8, 10)

Question 11.
The vertices of a triangle are A (-2, 2), B (4, 4) and C (8, 2). Find the length of the medians to the sides (i) AB, (ii) AC and (iii) BC.
Solution:
Let D, E and F be the mid-points of the sides
AB, BC and CA of the ∆ABC. AE, BF and CD are joined.
The vertices of ∆ABC are A (- 2, 2), B (4, 4) and C (8, 2)

∵ D is mid-point of AB
∴ Co-ordinates of D will be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) or \(\left(\frac{-2+4}{2}, \frac{2+4}{2}\right)\) or \(\left(\frac{2}{2}, \frac{6}{2}\right)\) or (1, 3)
Similarly co-ordinates of E will be \(\left(\frac{4+8}{2}, \frac{4+2}{2}\right)\) or (6, 3) and of F will be
\(\left(\frac{8-2}{2}, \frac{2+2}{2}\right)\) or (3, 2)
Now length of AE = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(-2-6)^2+(2-3)^2}=\sqrt{(-8)^2+(-1)^2}\)
= \(\sqrt{64+1}=\sqrt{65}\)
Similarly,
Length of BF = \(\sqrt{(4-3)^2+(4-2)^2}\)
= \(\sqrt{(1)^2+(2)^2}=\sqrt{1+4}=\sqrt{5}\)
and length of CD = \(\sqrt{(8-1)^2+(2-3)^2}\)
= \(\sqrt{(7)^2+(-1)^2}=\sqrt{49+1}=\sqrt{50}\)
= \(\sqrt{25 \times 2}=5 \sqrt{2}\)
∴ Length of median to AB, BC and CA are 5\(\sqrt{2}\), \(\sqrt{5}\) and \(\sqrt{65}\).

Question 12.
Calculate the coordinates of the point P which divides the line joining A (- 1,3), B (5, 9) in the ratio 1 : 2.
Solution:
Let the co-ordinates of the point P be (x, y) and m₁ : m₂ = 1 : 2

Question 13
The mid-point of the line joining A (2,p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Solution:
Let P (3, 5) be the mid-point of line joining the points A (2, p) and B (q, 4), then
∵ x = \(\frac{x_1+x_2}{2}\) and y = \(\frac{y_1+y_2}{2}\)
∴ 3 = \(\frac{2+q}{2}\) and 5 = \(\frac{p+4}{2}\)
⇒ 2 + q = 6 and p + 4 = 10
⇒ q = 6 – 2 = 4 and p = 10 – 4 = 6
∴ p = 6, 9 = 4

Question 14.
(a) A is the point on the y-axis whose ordinate is 5 and B is the point (-3, 1). Calculate the length of AB.
(b) The mid-point of the line joining (a, 2) and (3, 6) is (2, b). Find the numerical values of a and b.
Solution:
(a) ∵ Point A is on y-axis
∴ its abscissa i.e., x co-ordinates will be 0
∴ Co-ordinates of A will be (0, 5)
and co-ordinates of B are (-3, 1)
∴ Length of AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(-3-0)^2+(1-5)^2}\)
\(\sqrt{(-3)^2+(-4)^2}\) = 9 + 16 = 25 = (5)²
∴ AB = 5

(b) Let point P (2, b) be the mid-point of the line joining A (a, 2) and b (3, 6)
But x = \(\frac{x_1+x_2}{2} \text { and } y=\frac{y_1+y_2}{2}\)
∴ 2 = \(\frac{a+3}{2}\) ⇒ a + 3 = 4 ⇒ a = 4 – 3 = 1
and b = \(\frac{2+6}{2}\) ⇒ 2b = 8 ⇒ b = \(\frac { 8 }{ 2 }\) = 4
∴ a = 1, b = 4

Question 15.
The line joining A (2, 3) and B (6, – 5) meets the x-axis at P. Write down the y-coordinates of P. Hence find the ratio AP : PB.
Solution:
A (2, 3) and B (6, -5) are the points which form a line AB
∵ P lies on x-axis
∴ coordinates of P will be 0.
Let P divides AB in the ratio m₁ : m₂, then
y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2} \Rightarrow 0=\frac{m_1(-5)+m_2(3)}{m_1+m_2}\)
⇒ \(\frac{-5 m_1+3 m_2}{m_1+m_2}\) = 0 ⇒ – 5₁ + 3m₂ = 0
⇒ 3m₂ = 5m₁
⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
∴ Ratio = 3 : 5 or AP : PB = 3 : 5

Question 16.
In the given figure P (2, 3) is the mid-point of the line AB. Write down the coordinates of A and B.

Solution:
In the figure, point A lies on x-axis
and B lies on.y-axis
∴ y-coordinates of A will be 0
and x-coordinates B will be 0
Let the co-ordinates of A and B be (x, 0) and (0, y) respectively.
∴ P (2, 3) is the mid-point of AB.
∴ 2 = \(\frac{x_1+x_2}{2}=\frac{x+0}{2}=\frac{x}{2}\)
∴ x = 2 x 2 = 4
and 3 = \(\frac{y_1+y_2}{2}=\frac{0+y}{2}=\frac{y}{2}\)
∴ y = 2 x 3 = 6
∴ Co-ordinates of A and B will be (4, 0) and (0, 6)

Question 17.
The line segment joining A (2, 3) and B (6, -5) is intersected by the x-axis at a point K. Write down the ordinate of the point K. Hence find the ratio in which K divides AB.
Solution:
∵ The line joining the points A (2, 3) and B (6, -5) intersects x-axis at K
∴ Ordinate or y-coordinates of K will be 0
Let K divides AB in the ratio m₁ : m₂
∴ y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2} \Rightarrow 0=\frac{m_1(-5)+m_2(3)}{m_1+m_2}\)
⇒ \(\frac{-5 m_1+3 m_2}{m_1+m_2}\) = 0 ⇒ – 5m₁ + 3m₂ = 0
⇒ 3m₂ = 5m₁ ⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
∴ Ratio = 3 : 5

Question 18.
(a) Coordinates of A and B are (-3, a) and (1, a + 4). The mid-point of AB is (-1, 1). Find the value of a.
(b) Calculate the ratio in which the line segment joining (3., 4) and (-2, 1) is divided by they-axis.
Solution:
(a) Co-ordinates of A are (- 3, a) and of B (1, a + 4)
Let mid-point of AB is P (- 1, 1)
∴ y = \(\frac{y_1+y_2}{2}\) ⇒ 1 = \(\frac{a+a+4}{2}\) ⇒ 2a + 4 = 2
⇒ 2a = 2 – 4 ⇒ 2a = – 2
∴ a = \(\frac { -2 }{ 2 }\) = – 1

(b) Let a point P divides the line segment joining the points A (3, 4) and B (-2, 1) in the ratio w₁ : m₂
∵ P lies on the y-axis
∴ Its abscissa (x-coordiante) will be 0
∴ But x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\)
⇒ 0 = \(\frac{m_1(-2)+m_2(3)}{m_1+m_2}=\frac{-2 m_1+3 m_2}{m_1+m_2}\)
⇒ – 2m₁+ 3m₂ = 0 ⇒ 3m₂ = 2m₁.
⇒ \(\frac{m_1}{m_2}=\frac{3}{2}\)
∴ Ratio = 3:2

Question 19.
(a) P divides the distance between A (-2, 1) and B (1,4) in the ratio 2 : 1. Calculate the coordinates of the point P.
(b) Prove that the point A (-5, 4), B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the coordinates of D, so that ABCD is a square.
Solution:
(a) P divides the line segment AB whose vertices are A (-2, 1) and B (1, 4) in the ratio 2 : 1 Let co-ordinates of P be (x₁ y)
Here m₁ : m₂ = 2 : 1

Co-ordinates of P will be (0, 3)

(b) Co-ordinates of A are (-5, 4), of B are (-1, -2) and of C are (5, 2)

AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{[-1-(-5)]^2+(-2-4)^2}\)
= \(\sqrt{(-1+5)^2+(-6)^2}=\sqrt{(4)^2+(-6)^2}\)
= \(\sqrt{16+36}=\sqrt{52}\)
∴ AB² = 52
Similarly
BC² = [5 – (-1)]² + [2 – (-2)]²
= (5 + 1)² + (2 + 2)² = (6 )² + (4)² = 36 + 16 = 52
and AC² = [5 – (-5)]² + (2 – 4)² = (5 + 5)² + (2 – 4)²
= (10)² + (-2)² = 100 + 4 = 104
∵ AB = AC and
AB² + BC² = AC²
∴∆ABC is an isosceles right triangle whose ∠B = 90°
∵ ABCD is a square
Let co-ordinates of D be (a, b)
Join BD which intersects AC at O
∵ Diagonals of a square bisect each other
∴ O is mid-point of AC as well as BD
Now co-ordinates of O will be
\(\left(\frac{5-5}{2}, \frac{2+4}{2}\right) \text { or }\left(\frac{0}{2}, \frac{6}{2}\right)\) or (0, 3)
∵ O is mid-point of BD, then
0 = \(\frac{-1+a}{2}\) ⇒ – 1 + a = 0 ⇒ a = 1
and 3 = \(\frac{-2+b}{2}\) ⇒ – 2 + 6 = 6 ⇒ 6 = 6 + 2 = 8
∴ Co-ordinates of D are (1, 8)

Question 20.
A (2, 2), B (-2, 4), C (2, 6) are the vertices of a triangle ABC. Prove that ABC is an isosceles triangle.
Solution:
Vertices of a AABC are A (2, 2), B (-2, 4). C (2, 6)
Now AB² = (x₂ – x₁)² + (y₂ – y₁)²
= (-2 – 2)² + (4 – 2)² = (- 4)² + (2)²
= 16 + 4 = 20
BC² = [2 – (-2)]² + (6 – 4)² = (2 + 2)² + (2)²
= (4)² + (2)² = 16 + 4 = 20
AC² = (2 – 2)² + (2 – 6)²
= 0² + (-4)²
= 0 + 16 = 16
∵ AB² = BC² ⇒ AB = BC
∴ ∆ABC is an isosceles triangle.