| Exercise
10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
Exercise 10.2
Question 1 A
Find the distance between the following pair of points:
(0, 0), (- 5, 12)
Sol :(0, 0), (- 5, 12)
Given points are (0, 0) and (- 5, 12)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(0-(-5))^{2}+(0-12)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(0-(-5))^{2}+(0-12)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(5)^{2}+(-12)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{25+144}$
⇒ S = √169⇒ S = 13
∴ The distance between the points (0, 0) and (- 5, 12) is 13 units.
Question 1 B
Find the distance between the following pair of points:
(4, 5), (- 3, 2)
Sol :(4, 5), (- 3, 2)
Given points are (4, 5) and (- 3, 2)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
∴ The distance between the points (4, 5) and (- 3, 2) is √58 units.
Given points are (5, - 12) and (9, - 9)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\Rightarrow \mathrm{S}=\sqrt{(4-(-3))^{2}+(5-2)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(7)^{2}+(3)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{49+9}$
⇒ S = √58∴ The distance between the points (4, 5) and (- 3, 2) is √58 units.
Question 1 C
Find the distance between the following pair of points:
(5, - 12), (9, - 9)
Sol :(5, - 12), (9, - 9)
Given points are (5, - 12) and (9, - 9)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ S = √25
⇒ S = 5
∴ The distance between the points (5, -12) and (9, - 9) is 5 units.
Given points are (- 3, 4) and (3, 0)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is \
$\Rightarrow \mathrm{S}=\sqrt{(5-9)^{2}+(-12-(-9))^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(-4)^{2}+(-3)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{16+9}$
⇒ S = √25
⇒ S = 5
∴ The distance between the points (5, -12) and (9, - 9) is 5 units.
Question 1 D
Find the distance between the following pair of points:
(- 3, 4), (3, 0)
Sol :(- 3, 4), (3, 0)
Given points are (- 3, 4) and (3, 0)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is \
$\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}$
⇒ S = √52
⇒ $S=\sqrt{4 \times 13}$
⇒ S = 2√13
∴ The distance between the points (- 3, 4) and (3, 0) is 2√13 units.
Given points are (2, 3) and (4, 1)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\Rightarrow \mathrm{S}=\sqrt{(-3-3)^{2}+(4-0)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(-6)^{2}+(4)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{36+16}$
⇒ S = √52
⇒ $S=\sqrt{4 \times 13}$
⇒ S = 2√13
∴ The distance between the points (- 3, 4) and (3, 0) is 2√13 units.
Question 1 E
Find the distance between the following pair of points:
(2, 3), (4, 1)
Sol :(2, 3), (4, 1)
Given points are (2, 3) and (4, 1)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(2-4)^{2}+(3-1)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(-2)^{2}+(2)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{4+4}$
⇒ S = √8
⇒ $s=\sqrt{4 \times 2}$
⇒ S = 2√2
∴ The distance between the points (2, 3) and (4, 1) is 2√2 units.
Question 1 F
Find the distance between the following pair of points:
(a, b), (- a, - b)
Sol :(a, b), (- a, - b)
Given points are (a, b) and (- a, - b)
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Given that we need to show that the points (1, - 1), (- 5, 7) and (2, 6) are equidistant from the point (- 2, 3).
We know that distance between two points (x1, y1) and (x2, y2) is
$\Rightarrow
\mathrm{S}=\sqrt{(\mathrm{a}-(-\mathrm{a}))^{2}+(\mathrm{b}-(-\mathrm{b}))^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(2 \mathrm{a})^{2}+(2 \mathrm{b})^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{4 \mathrm{a}^{2}+4 \mathrm{b}^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{4 \times\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)}$
$\Rightarrow \mathrm{S}=2 \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$
∴ The distance between the points (a, b) and (- a, - b) is $2 \sqrt{a^{2}+b^{2}}$ unitsQuestion 2
Examine whether the points (1, - 1), (- 5, 7) and (2, 6) are equidistant from the point (- 2, 3)?
Sol :Given that we need to show that the points (1, - 1), (- 5, 7) and (2, 6) are equidistant from the point (- 2, 3).
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Let S1 be the distance between the points (1, - 1) and (- 2, 3)
$\Rightarrow \mathrm{S}_{1}=\sqrt{(1-(-2))^{2}+(-1-3)^{2}}$
$\Rightarrow \mathrm{S}_{1}=\sqrt{(3)^{2}+(-4)^{2}}$
$\Rightarrow \mathrm{S}_{1}=\sqrt{9+16}$
$\Rightarrow \mathrm{S}_{1}=\sqrt{25}$
⇒ S1 = 5 ..... (1)Let S2 be the distance between the points (- 5, 7) and (- 2, 3)
$\Rightarrow \mathrm{S}_{2}=\sqrt{(-5-(-2))^{2}+(7-3)^{2}}$
$\Rightarrow \mathrm{S}_{2}=\sqrt{(-3)^{2}+(4)^{2}}$
$\Rightarrow \mathrm{S}_{2}=\sqrt{9+16}$
$\Rightarrow \mathrm{S}_{2}=\sqrt{25}$
⇒ S2 = 5 ..... (2)Let S3 be the distance between the points (2, 6) and (- 2, 3)
$\Rightarrow \mathrm{S}_{3}=\sqrt{(2-(-2))^{2}+(6-3)^{2}}$
$\Rightarrow \mathrm{S}_{3}=\sqrt{(4)^{2}+(3)^{2}}$
$\Rightarrow \mathrm{S}_{3}=\sqrt{16+9}$
$\Rightarrow \mathrm{S}_{3}=\sqrt{25}$
⇒ S3 = 5 ..... (3)From (1), (2), and (3) we got S1 = S2 = S3 which tells us that (1, - 1), (5, 7) and (2, 5) are equidistant from (- 2, 3).
Question 3 A
Find a if the distance between (a, 2) and (3, 4) is 8.
Sol :Given that the distance between the points (a, 2) and (3, 4) is 8.
We need to find the value of a.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$\Rightarrow 8=\sqrt{(a-3)^{2}+(2-4)^{2}}$
⇒ 82 = (a - 3)2 + (- 2)2
⇒ 64 = (a - 3)2 + 4
⇒ (a - 3)2 = 60
⇒ a – 3 = ± √60
⇒ a = 3 ± √60
∴ The values of a are 3±√60.
Given that the line has length of 10 units and one of its ends is (- 2, 3).
It is also given that the ordinate of the other end is 9. Let us assume the other end is (x, 9).
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ 82 = (a - 3)2 + (- 2)2
⇒ 64 = (a - 3)2 + 4
⇒ (a - 3)2 = 60
⇒ a – 3 = ± √60
⇒ a = 3 ± √60
∴ The values of a are 3±√60.
Question 3 B
A line is of length 10 units and one of its ends is (- 2, 3). If the ordinate of the other end is 9, prove
that the absicca of the other end is 6 or - 10.
Sol :Given that the line has length of 10 units and one of its ends is (- 2, 3).
It is also given that the ordinate of the other end is 9. Let us assume the other end is (x, 9).
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}$
⇒ $10=\sqrt{(-2-x)^{2}+(3-9)^{2}}$
⇒ 10 = (x + 2)2 + (- 6)2
⇒ 100 = (x + 2)2 + 36
⇒ (x + 2)2 = 64
⇒ x + 2 = ±8
⇒ x = - 2 + 8 (or) x = - 2 - 8
⇒ x = 6 or 10
∴ Thus proved.
Given that the distance between the points P(2, - 3) and Q(10, y) is 10.
We need to find the value of y.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ $10=\sqrt{(-2-x)^{2}+(3-9)^{2}}$
⇒ 10 = (x + 2)2 + (- 6)2
⇒ 100 = (x + 2)2 + 36
⇒ (x + 2)2 = 64
⇒ x + 2 = ±8
⇒ x = - 2 + 8 (or) x = - 2 - 8
⇒ x = 6 or 10
∴ Thus proved.
Question 3 C
Find the value of y for which the distance between the points P(2, - 3) and Q(10, y) is 10 units.
Sol :Given that the distance between the points P(2, - 3) and Q(10, y) is 10.
We need to find the value of y.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ $10=\sqrt{(2-10)^{2}+(-3-\mathrm{y})^{2}}$
⇒ 102 = (- 8)2 + (3 + y)2
⇒ 100 = 64 + (3 + y)2
⇒ (3 + y)2 = 36
⇒ 3 + y = ±6
⇒ y = 3 + 6 (or) y = 3 - 6
⇒ y = 9 (or) y = - 3
∴ The values of y are 9, - 3.
Given points are (at12, 2at1) and (at22, 2at2).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
⇒ $10=\sqrt{(2-10)^{2}+(-3-\mathrm{y})^{2}}$
⇒ 102 = (- 8)2 + (3 + y)2
⇒ 100 = 64 + (3 + y)2
⇒ (3 + y)2 = 36
⇒ 3 + y = ±6
⇒ y = 3 + 6 (or) y = 3 - 6
⇒ y = 9 (or) y = - 3
∴ The values of y are 9, - 3.
Question 4 A
Find the distance between the points:
(at12, 2at1) and (at22, 2at2)
Sol :(at12, 2at1) and (at22, 2at2)
Given points are (at12, 2at1) and (at22, 2at2).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Given points are (a - b, b - a) and (a + b, a + b).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\Rightarrow
\mathrm{S}=\sqrt{\left(\mathrm{at}_{1}^{2}-\mathrm{at}_{2}^{2}\right)^{2}+\left(2 \mathrm{at}_{1}-2
\mathrm{at}_{2}\right)^{2}}$
$\Rightarrow
\mathrm{S}=\sqrt{\left(\mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)\right)^{2}+\left(2
\mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)\right)^{2}}$
$\Rightarrow \mathrm{S}=\left(\mathrm{a}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)\right)
\sqrt{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}+2^{2}}$
$\Rightarrow \mathrm{S}=\left(\mathrm{a}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)\right)
\sqrt{\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}+4}$
∴ The distance between the points (at12, 2at1) and (at22,
2at2) is $\left(a\left(t_{1}+t_{2}\right)\right) \sqrt{\left(t_{1}+t_{2}\right)^{2}+4}$Question 4 B
Find the distance between the points:
(a - b, b - a) and (a + b, a + b)
Sol :(a - b, b - a) and (a + b, a + b)
Given points are (a - b, b - a) and (a + b, a + b).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$\Rightarrow
\mathrm{S}=\sqrt{((\mathrm{a}-\mathrm{b})-(\mathrm{a}+\mathrm{b}))^{2}+((\mathrm{b}-\mathrm{a})-(\mathrm{a}+\mathrm{b}))^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{(-2 \mathrm{b})^{2}+(-2 \mathrm{a})^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{4\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)}$
$\Rightarrow \mathrm{S}=2 \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$
∴ The distance between the points (a - b, b - a) and (a + b, a + b) is $2 \sqrt{a^{2}+b^{2}}$Question 4 C
Find the distance between the points:
(cosθ, sinθ) and (sinθ, cosθ)
Sol :(cosθ, sinθ) and (sinθ, cosθ)
Given points are (cosθ, sinθ) and (sinθ, cosθ).
We need to find the distance between these two points.
We know that distance(S) between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Given points are A(7, 6) and B(- 3, 4).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is
$\Rightarrow \mathrm{S}=\sqrt{(\cos \theta-\sin \theta)^{2}+(\sin \theta-\cos \theta)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{\cos ^{2} \theta+\sin ^{2} \theta-2 \sin \theta \cos
\theta+\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \theta \cos \theta}$
$\Rightarrow \mathrm{S}=\sqrt{2\left(\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \theta \cos
\theta\right)}$
$\Rightarrow \mathrm{S}=\sqrt{2(\sin \theta-\cos \theta)^{2}}$
$\Rightarrow \mathrm{S}=\sqrt{2}(\sin \theta-\cos \theta)$
∴ The distance between the points (cosθ, sinθ) and (sinθ, cosθ) is $\sqrt{2}(\sin \theta-\cos \theta)$Question 5 A
Find the point on x - axis which is equidistant from the following pair of points:
(7, 6) and (- 3, 4)
Sol :(7, 6) and (- 3, 4)
Given points are A(7, 6) and B(- 3, 4).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 7)2 + (0 - 6)2 = (x - (- 3))2 + (0 - 4)2
⇒ (x - 7)2 + (- 6)2 = (x + 3)2 + (- 4)2
⇒ x2 - 14x + 49 + 36 = x2 + 6x + 9 + 16
⇒ 20x = 60
$\Rightarrow x=\frac{60}{20}$
⇒ x = 3
∴ The point on x - axis is (3, 0).
Given points are A(3, 2) and B(- 5, - 2).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 7)2 + (0 - 6)2 = (x - (- 3))2 + (0 - 4)2
⇒ (x - 7)2 + (- 6)2 = (x + 3)2 + (- 4)2
⇒ x2 - 14x + 49 + 36 = x2 + 6x + 9 + 16
⇒ 20x = 60
$\Rightarrow x=\frac{60}{20}$
⇒ x = 3
∴ The point on x - axis is (3, 0).
Question 5 B
Find the point on x - axis which is equidistant from the following pair of points:
(3, 2) and (- 5, - 2)
Sol :(3, 2) and (- 5, - 2)
Given points are A(3, 2) and B(- 5, - 2).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 3)2 + (0 - 2)2 = (x + 5)2 + (0 - (- 2))2
⇒ (x - 3)2 + (- 2)2 = (x + 5)2 + (2)2
⇒ x2 - 6x + 9 + 4 = x2 + 10x + 25 + 4
⇒ 16x = - 16
$\Rightarrow x=\frac{-16}{16}$
⇒ x = - 1
∴ The point on x - axis is (- 1, 0).
Given points are A(2, - 5) and B(- 2, 9).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 2)2 + (0 - (- 5))2 = (x - (- 2))2 + (0 - 9)2
⇒ (x - 2)2 + (5)2 = (x + 2)2 + (- 9)2
⇒ x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
⇒ 8x = - 56
$\Rightarrow x=\frac{-56}{8}$
⇒ x = - 7
∴ The point on x - axis is (- 7, 0).
Given points are A(- 5, - 2) and B(3, 2).
We need to find a point on y - axis which is equidistant from these points.
Let us assume the point on y - axis be S(0, y).
We know that distance between the points (x1, y1) and (x2, y2) is
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 3)2 + (0 - 2)2 = (x + 5)2 + (0 - (- 2))2
⇒ (x - 3)2 + (- 2)2 = (x + 5)2 + (2)2
⇒ x2 - 6x + 9 + 4 = x2 + 10x + 25 + 4
⇒ 16x = - 16
$\Rightarrow x=\frac{-16}{16}$
⇒ x = - 1
∴ The point on x - axis is (- 1, 0).
Question 5 C
Find the point on x - axis which is equidistant from the following pair of points:
(2, - 5) and (- 2, 9)
Sol :(2, - 5) and (- 2, 9)
Given points are A(2, - 5) and B(- 2, 9).
We need to find a point on x - axis which is equidistant from these points.
Let us assume the point on x - axis be S(x, o).
We know that distance between the points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 2)2 + (0 - (- 5))2 = (x - (- 2))2 + (0 - 9)2
⇒ (x - 2)2 + (5)2 = (x + 2)2 + (- 9)2
⇒ x2 - 4x + 4 + 25 = x2 + 4x + 4 + 81
⇒ 8x = - 56
$\Rightarrow x=\frac{-56}{8}$
⇒ x = - 7
∴ The point on x - axis is (- 7, 0).
Question 6 A
Find the point on y - axis which is equidistant from point (- 5, - 2) and (3, 2).
Sol :Given points are A(- 5, - 2) and B(3, 2).
We need to find a point on y - axis which is equidistant from these points.
Let us assume the point on y - axis be S(0, y).
We know that distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (0 - (- 5))2 + (y - (- 2))2 = (0 - 3)2 + (y - 2)2
⇒ (5)2 + (y + 2)2 = (- 3)2 + (y - 2)2
⇒ 25 + y2 + 4y + 4 = 9 + y2 - 4y + 4
⇒ 8y = - 16
$\Rightarrow \mathrm{y}=\frac{-16}{8}$
⇒ y = - 2
∴ The point on y - axis is (0, - 2).
Given points are A(6, 5) and B(- 4, 3).
We need to find a point on y - axis which is equidistant from these points.
Let us assume the point on y - axis be S(0, y).
We know that distance between the points (x1, y1) and (x2, y2) is
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (0 - (- 5))2 + (y - (- 2))2 = (0 - 3)2 + (y - 2)2
⇒ (5)2 + (y + 2)2 = (- 3)2 + (y - 2)2
⇒ 25 + y2 + 4y + 4 = 9 + y2 - 4y + 4
⇒ 8y = - 16
$\Rightarrow \mathrm{y}=\frac{-16}{8}$
⇒ y = - 2
∴ The point on y - axis is (0, - 2).
Question 6 B
Find the point on y - axis which is equidistant from the points A(6, 5) and B(- 4, 3).
Sol :Given points are A(6, 5) and B(- 4, 3).
We need to find a point on y - axis which is equidistant from these points.
Let us assume the point on y - axis be S(0, y).
We know that distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (0 - 6)2 + (y - 5)2 = (0 - (- 4))2 + (y - 3)2
⇒ (- 6)2 + (y - 5)2 = (4)2 + (y - 3)2
⇒ 36 + y2 - 10y + 25 = 16 + y2 - 6y + 9
⇒ 4y = 36
$\Rightarrow \mathrm{y}=\frac{36}{4}$
⇒ y = 9
∴ The point on y - axis is (0, 9).
Given points are A(3, 5), B(1, 1) and C(- 2, - 5).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
From the problem,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (0 - 6)2 + (y - 5)2 = (0 - (- 4))2 + (y - 3)2
⇒ (- 6)2 + (y - 5)2 = (4)2 + (y - 3)2
⇒ 36 + y2 - 10y + 25 = 16 + y2 - 6y + 9
⇒ 4y = 36
$\Rightarrow \mathrm{y}=\frac{36}{4}$
⇒ y = 9
∴ The point on y - axis is (0, 9).
Question 7 A
Using distance formula, examine whether the following sets of points are collinear?
(3, 5), (1, 1), (- 2, - 5)
Sol :(3, 5), (1, 1), (- 2, - 5)
Given points are A(3, 5), B(1, 1) and C(- 2, - 5).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ AC = 5√5 ..... (1)
⇒ AB = 2√5 ..... (2)
⇒ BC = 3√5 ..... (3)
From (1), (2), (3) we can see that AB + BC = AC.
∴ The three points are collinear.
Given points are A(5, 1), B(1, - 1) and C(11, 4).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\Rightarrow A C=\sqrt{(3-(-2))^{2}+(5-(-5))^{2}}$
$\Rightarrow A C=\sqrt{(5)^{2}+(10)^{2}}$
$\Rightarrow A C=\sqrt{25+100}$
$\Rightarrow A C=\sqrt{125}$
$\Rightarrow A C=\sqrt{5 \times 25}$
⇒ AC = 5√5 ..... (1)
$\Rightarrow \mathrm{AB}=\sqrt{(3-1)^{2}+(5-1)^{2}}$
$\Rightarrow \mathrm{AB}=\sqrt{(2)^{2}+(4)^{2}}$
$\Rightarrow \mathrm{AB}=\sqrt{4+16}$
$\Rightarrow \mathrm{AB}=\sqrt{20}$
$\Rightarrow \mathrm{AB}=\sqrt{4 \times 5}$
$\Rightarrow B C=\sqrt{(1-(-2))^{2}+(1-(-5))^{2}}$
$\Rightarrow B C=\sqrt{(3)^{2}+(6)^{2}}$
$\Rightarrow B C=\sqrt{9+36}$
$\Rightarrow B C=\sqrt{45}$
$\Rightarrow B C=\sqrt{9 \times 5}$
⇒ BC = 3√5 ..... (3)
From (1), (2), (3) we can see that AB + BC = AC.
∴ The three points are collinear.
Question 7 B
Using distance formula, examine whether the following sets of points are collinear?
(5, 1), (1, - 1), (11, 4)
Sol :(5, 1), (1, - 1), (11, 4)
Given points are A(5, 1), B(1, - 1) and C(11, 4).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ AC = 3√5 ..... (1)
⇒ AB = 2√5 ..... (2)
⇒ BC = 5√5 ..... (3)
From (1), (2), (3) we can see that AB + AC = BC.
∴ The three points are collinear.
Given points are A(0, 0), B(9, 6) and C(3, 2).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\Rightarrow A C=\sqrt{(5-11)^{2}+(1-4)^{2}}$
$\Rightarrow A C=\sqrt{(-6)^{2}+(-3)^{2}}$
$\Rightarrow A C=\sqrt{36+9}$
$\Rightarrow A C=\sqrt{45}$
$\Rightarrow A C=\sqrt{9 \times 5}$
⇒ AC = 3√5 ..... (1)
$\Rightarrow A B=\sqrt{(5-1)^{2}+(1-(-1))^{2}}$
$\Rightarrow A B=\sqrt{(4)^{2}+(2)^{2}}$
$\Rightarrow A B=\sqrt{16+4}$
$\Rightarrow A B=\sqrt{20}$
$\Rightarrow A B=\sqrt{4 \times 5}$
⇒ AB = 2√5 ..... (2)
$\Rightarrow B C=\sqrt{(1-11)^{2}+(-1-4)^{2}}$
$\Rightarrow B C=\sqrt{(-10)^{2}+(-5)^{2}}$
$\Rightarrow B C=\sqrt{100+25}$
$\Rightarrow B C=\sqrt{125}$
$\Rightarrow B C=\sqrt{25 \times 5}$
⇒ BC = 5√5 ..... (3)
From (1), (2), (3) we can see that AB + AC = BC.
∴ The three points are collinear.
Question 7 C
Using distance formula, examine whether the following sets of points are collinear?
(0, 0), (9, 6), (3, 2)
Sol :(0, 0), (9, 6), (3, 2)
Given points are A(0, 0), B(9, 6) and C(3, 2).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ AC = √13 ..... (1)
⇒ AB = 3√13 ..... (2)
⇒ BC = 2√13 ..... (3)
From (1), (2), (3) we can see that AB = BC + AC.
∴ The three points are collinear.
Given points are A(- 1, 2), B(5, 0) and C(2, 1).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\Rightarrow A C=\sqrt{(0-3)^{2}+(0-2)^{2}}$
$\Rightarrow A C=\sqrt{(-3)^{2}+(-2)^{2}}$
$\Rightarrow A C=\sqrt{9+4}$
⇒ AC = √13 ..... (1)
$\Rightarrow A B=\sqrt{(0-9)^{2}+(0-6)^{2}}$
$\Rightarrow A B=\sqrt{(-9)^{2}+(-6)^{2}}$
$\Rightarrow A B=\sqrt{81+36}$
$\Rightarrow A B=\sqrt{117}$
$\Rightarrow A B=\sqrt{9 \times 13}$
⇒ AB = 3√13 ..... (2)
$\Rightarrow B C=\sqrt{(9-3)^{2}+(6-2)^{2}}$
$\Rightarrow B C=\sqrt{(6)^{2}+(4)^{2}}$
$\Rightarrow B C=\sqrt{36+16}$
$\Rightarrow B C=\sqrt{52}$
$\Rightarrow B C=\sqrt{4 \times 13}$
⇒ BC = 2√13 ..... (3)
From (1), (2), (3) we can see that AB = BC + AC.
∴ The three points are collinear.
Question 7 D
Using distance formula, examine whether the following sets of points are collinear?
(- 1, 2), (5, 0), (2, 1)
Sol :(- 1, 2), (5, 0), (2, 1)
Given points are A(- 1, 2), B(5, 0) and C(2, 1).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is a linear relationship between AB, BC and AC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ AC = √10 ..... (1)
⇒ AB = 2√10 ..... (2)
⇒ BC = √10 ..... (3)
From (1), (2), (3) we can see that AC + BC = AB.
∴ The three points are collinear.
Given points are A(1, 5), B(2, 3) and C(- 2, - 11).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\Rightarrow A C=\sqrt{(-1-2)^{2}+(2-1)^{2}}$
$\Rightarrow A C=\sqrt{(-3)^{2}+(1)^{2}}$
$\Rightarrow A C=\sqrt{9+1}$
⇒ AC = √10 ..... (1)
$\Rightarrow A B=\sqrt{(-1-5)^{2}+(2-0)^{2}}$
$\Rightarrow A B=\sqrt{(-6)^{2}+(2)^{2}}$
$\Rightarrow A B=\sqrt{36+4}$
$\Rightarrow A B=\sqrt{40}$
$\Rightarrow A B=\sqrt{4 \times 10}$
⇒ AB = 2√10 ..... (2)
$\Rightarrow B C=\sqrt{(5-2)^{2}+(0-1)^{2}}$
$\Rightarrow B C=\sqrt{(3)^{2}+(-1)^{2}}$
$\Rightarrow B C=\sqrt{9+1}$
⇒ BC = √10 ..... (3)
From (1), (2), (3) we can see that AC + BC = AB.
∴ The three points are collinear.
Question 7 E
Using distance formula, examine whether the following sets of points are collinear?
(1, 5), (2, 3), (- 2, - 11)
Sol :(1, 5), (2, 3), (- 2, - 11)
Given points are A(1, 5), B(2, 3) and C(- 2, - 11).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AC = AB + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ AC = √265 ..... (1)
⇒ AB = √5 ..... (2)
⇒ BC = 2√53 ..... (3)
From (1), (2), (3) we can see that we cannot get any linear relationship.
∴ The three points are not collinear.
Given points are A(6, 1), B(1, 3) and C(x, 8). We need to find the value of x such that AB = BC.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒ AB = BC
⇒ AB2 = BC2
⇒ (6 - 1)2 + (1 - 3)2 = (1 - x)2 + (3 - 8)2
⇒ (5)2 + (- 2)2 = (1 - x)2 + (- 5)2
⇒ 25 + 4 = 1 - 2x + x2 + 25
⇒ x2 - 2x - 3 = 0
⇒ x2 - 3x + x - 3 = 0
⇒ x(x - 3) + 1(x - 3) = 0
⇒ (x + 1)(x - 3) = 0
⇒ x + 1 = 0 (or) x - 3 = 0
⇒ x = - 1 (or) x = 3
∴ The values of x are - 1 or 3.
Given points are A(a + rcosθ, b + rsinθ) and B(a, b).
We know that the distance between the points (x1, y1) and (x2, y2) is
$\Rightarrow A C=\sqrt{(1-(-2))^{2}+(5-(-11))^{2}}$
$\Rightarrow A C=\sqrt{(3)^{2}+(16)^{2}}$
$\Rightarrow A C=\sqrt{9+256}$
⇒ AC = √265 ..... (1)
$\Rightarrow A B=\sqrt{(1-2)^{2}+(5-3)^{2}}$
$\Rightarrow A B=\sqrt{(-1)^{2}+(2)^{2}}$
$\Rightarrow A B=\sqrt{1+4}$
⇒ AB = √5 ..... (2)
$\Rightarrow B C=\sqrt{(2-(-2))^{2}+(3-(-11))^{2}}$
$\Rightarrow B C=\sqrt{(4)^{2}+(14)^{2}}$
$\Rightarrow B C=\sqrt{16+196}$
$\Rightarrow B C=\sqrt{212}$
$\Rightarrow B C=\sqrt{4 \times 53}$
⇒ BC = 2√53 ..... (3)
From (1), (2), (3) we can see that we cannot get any linear relationship.
∴ The three points are not collinear.
Question 8
If A = (6, 1), B = (1, 3) and C = (x, 8), find the value of x such that AB = BC.
Sol :Given points are A(6, 1), B(1, 3) and C(x, 8). We need to find the value of x such that AB = BC.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒ AB = BC
⇒ AB2 = BC2
⇒ (6 - 1)2 + (1 - 3)2 = (1 - x)2 + (3 - 8)2
⇒ (5)2 + (- 2)2 = (1 - x)2 + (- 5)2
⇒ 25 + 4 = 1 - 2x + x2 + 25
⇒ x2 - 2x - 3 = 0
⇒ x2 - 3x + x - 3 = 0
⇒ x(x - 3) + 1(x - 3) = 0
⇒ (x + 1)(x - 3) = 0
⇒ x + 1 = 0 (or) x - 3 = 0
⇒ x = - 1 (or) x = 3
∴ The values of x are - 1 or 3.
Question 9
Prove that the distance between the points (a + rcosθ, b + rsinθ) and (a, b) is independent of θ.
Sol :Given points are A(a + rcosθ, b + rsinθ) and B(a, b).
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ AB = r
We can see that AB is independent of θ.
∴ Thus proved.
Given points are A(cosec2θ, 0), B(0, sec2θ ) and C(1, 1).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AB = AC + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\Rightarrow A B=\sqrt{(a+r \cos \theta-a)^{2}+(b+r \sin \theta-b)^{2}}$
$\Rightarrow A B=\sqrt{(r \cos \theta)^{2}+(r \sin \theta)^{2}}$
$\Rightarrow A B=\sqrt{r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta}$
$\Rightarrow A B=\sqrt{r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}$
$\Rightarrow A B=\sqrt{r^{2}(1)}$
⇒ AB = r
We can see that AB is independent of θ.
∴ Thus proved.
Question 10 A
use distance formula to show that the points (cosec2θ, 0), (0, sec2θ) and (1, 1) are
collinear.
Sol :Given points are A(cosec2θ, 0), B(0, sec2θ ) and C(1, 1).
We need to check whether these points are collinear.
We know that for three points A, B and C to be collinear, the criteria to be satisfied is AB = AC + BC.
Let us find the distances first,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$\Rightarrow A C=\frac{\sqrt{1+\tan ^{4} \theta}}{\tan ^{2} \theta}$ ..... (1)
$\Rightarrow \mathrm{AB}=\left(\frac{\sec ^{2} \theta}{\tan ^{2} \theta}\right) \sqrt{\left(1+\tan ^{4} \theta\right)}$ .... - (2)
⇒ $\mathrm{BC}=\sqrt{1+\tan ^{4} \theta}$ ..... (3)
Now,
$\Rightarrow A C+B C=\left(\frac{1}{\tan ^{2} \theta}\right) \sqrt{\left(1+\tan ^{4} \theta\right)}+\sqrt{1+\tan ^{4} \theta}$
⇒ AC + BC = AB
∴ The three points are collinear.
Given that circle passes through the points A(6, 2), B(0, 4), C(4, 6).
Let us assume O(x, y) be the centre of the circle.
We know that distance from the centre to any point on h circle is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
$\Rightarrow A C=\sqrt{\left(\operatorname{cosec}^{2} \theta-1\right)^{2}+(0-1)^{2}}$
$\Rightarrow A C=\sqrt{\left(\cot ^{2} \theta\right)^{2}+1^{2}}$
$\Rightarrow A C=\sqrt{\cot ^{4} \theta+1}$
$\Rightarrow A C=\sqrt{\frac{1}{\tan ^{4} \theta}+1}$
$\Rightarrow A C=\sqrt{\frac{1+\tan ^{4} \theta}{\tan ^{4} \theta}}$
$\Rightarrow A C=\frac{\sqrt{1+\tan ^{4} \theta}}{\tan ^{2} \theta}$ ..... (1)
$\Rightarrow A B=\sqrt{\left(\operatorname{cosec}^{2} \theta-0\right)^{2}+\left(0-\sec ^{2}
\theta\right)^{2}}$
$\Rightarrow A B=\sqrt{\left(1+\cot ^{2} \theta\right)^{2}+\left(1+\tan ^{2}
\theta\right)^{2}}$
$\Rightarrow A B=\sqrt{\left(1+\frac{1}{\tan ^{2} \theta}\right)^{2}+\left(1+\tan ^{2}
\theta\right)^{2}}$
$\Rightarrow A B=\left(1+\tan ^{2} \theta\right) \sqrt{\left(\frac{1+\tan ^{4} \theta}{\tan
^{4} \theta}\right)}$
$\Rightarrow \mathrm{AB}=\left(\frac{\sec ^{2} \theta}{\tan ^{2} \theta}\right) \sqrt{\left(1+\tan ^{4} \theta\right)}$ .... - (2)
$\Rightarrow B C=\sqrt{(0-1)^{2}+\left(\sec ^{2} \theta-1\right)^{2}}$
$\Rightarrow B C=\sqrt{1+\left(\tan ^{2} \theta\right)^{2}}$
⇒ $\mathrm{BC}=\sqrt{1+\tan ^{4} \theta}$ ..... (3)
Now,
$\Rightarrow A C+B C=\left(\frac{1}{\tan ^{2} \theta}\right) \sqrt{\left(1+\tan ^{4} \theta\right)}+\sqrt{1+\tan ^{4} \theta}$
$\Rightarrow A C+B C=\sqrt{1+\tan ^{4} \theta}\left(\frac{1}{\tan ^{2} \theta}+1\right)$
$\Rightarrow A C+B C=\sqrt{1+\tan ^{4} \theta}\left(\frac{1+\tan ^{2} \theta}{\tan ^{2}
\theta}\right)$
$\Rightarrow A C+B C=\sqrt{1+\tan ^{4} \theta}\left(\frac{\sec ^{2} \theta}{\tan ^{2}
\theta}\right)$
⇒ AC + BC = AB
∴ The three points are collinear.
Question 10 B
Using distance formula show that (3, 3) is the centre of the circle passing through the points (6, 2), (0,
4) and (4, 6). Find the radius of the circle.
Sol :Given that circle passes through the points A(6, 2), B(0, 4), C(4, 6).
Let us assume O(x, y) be the centre of the circle.
We know that distance from the centre to any point on h circle is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}$
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - 6)2 + (y - 2)2 = (x - 0)2 + (y - 4)2
⇒ x2 - 12x + 36 + y2 - 4y + 4 = x2 + y2 - 8y + 16
⇒ 12x - 4y = 24
⇒ 3x - y = 6 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 0)2 + (y - 4)2 = (x - 4)2 + (y - 6)2
⇒ x2 + y2 - 8y + 16 = x2 - 8x + 16 + y2 - 12y + 36
⇒ 8x + 4y = 36
⇒ 2x + y = 9 .... - (2)
On solving (1) and (2), we get
⇒ x = 3 and y = 3
∴ (3, 3) is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the radius of the circle.
⇒ r = √10
∴ The radius of the circle is √10.
Given points are A(2, 3) and B(6, - 1). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - 6)2 + (y - 2)2 = (x - 0)2 + (y - 4)2
⇒ x2 - 12x + 36 + y2 - 4y + 4 = x2 + y2 - 8y + 16
⇒ 12x - 4y = 24
⇒ 3x - y = 6 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 0)2 + (y - 4)2 = (x - 4)2 + (y - 6)2
⇒ x2 + y2 - 8y + 16 = x2 - 8x + 16 + y2 - 12y + 36
⇒ 8x + 4y = 36
⇒ 2x + y = 9 .... - (2)
On solving (1) and (2), we get
⇒ x = 3 and y = 3
∴ (3, 3) is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the radius of the circle.
$\Rightarrow \mathrm{r}=\mathrm{OA}=\sqrt{(3-6)^{2}+(3-2)^{2}}$
$\Rightarrow \mathrm{r}=\sqrt{(-3)^{2}+(1)^{2}}$
$\Rightarrow \mathrm{r}=\sqrt{9+1}$
⇒ r = √10
∴ The radius of the circle is √10.
Question 11 A
If the point (x, y) on the tangent is equidistant from the points (2, 3) and (6, - 1), find the relation
between x and y.
Sol :Given points are A(2, 3) and B(6, - 1). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y - (- 1))2
⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y + 1)2
⇒ x2 - 4x + 4 + y2 - 6y + 9 = x2 - 12x + 36 + y2 + 2y + 1
⇒ 8x - 8y = 24
⇒ x - y = 3
∴ The relation between x and y is x - y = 3.
Given points are A(7, 1) and B(3, 5). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y - (- 1))2
⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y + 1)2
⇒ x2 - 4x + 4 + y2 - 6y + 9 = x2 - 12x + 36 + y2 + 2y + 1
⇒ 8x - 8y = 24
⇒ x - y = 3
∴ The relation between x and y is x - y = 3.
Question 11 B
Find a relation between x and y such that the point (x, y) is equidistant from points (7, 1) and (3,
5).
Sol :Given points are A(7, 1) and B(3, 5). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 7)2 + (y - 1)2 = (x - 3)2 + (y - 5)2
⇒ x2 - 14x + 49 + y2 - 2y + 1 = x2 - 6x + 9 + y2 - 10y + 25
⇒ 8x - 8y = 16
⇒ x - y = 2
∴ The relation between x and y is x - y = 2.
Question 12 A
If the distances of P(x, y) from points A(3, 6) and B(- 3, 4) are equal, prove that 3x + y = 5
Sol :Given points are A(3, 6) and B(- 3, 4). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 3)2 + (y - 6)2 = (x - (- 3))2 + (y - 4)2
⇒ (x - 3)2 + (y - 6)2 = (x + 3)2 + (y - 4)2
⇒ x2 - 6x + 9 + y2 - 12y + 36 = x2 + 6x + 9 + y2 - 8y + 16
⇒ 12x + 4y = 20
⇒ 3x + y = 5
∴ Thus proved.
Given points are A(a + b, b - a) and B(a - b, a + b). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - 3)2 + (y - 6)2 = (x - (- 3))2 + (y - 4)2
⇒ (x - 3)2 + (y - 6)2 = (x + 3)2 + (y - 4)2
⇒ x2 - 6x + 9 + y2 - 12y + 36 = x2 + 6x + 9 + y2 - 8y + 16
⇒ 12x + 4y = 20
⇒ 3x + y = 5
∴ Thus proved.
Question 12 B
If the point (x, y) be equidistant from the points (a + b, b - a) and (a - b, a + b), prove
that $\frac{a-b}{a+b}=\frac{x-y}{x+y}$
Sol :Given points are A(a + b, b - a) and B(a - b, a + b). It is told that S(x, y) is equidistant from A and B.
So, we get SA = SB,
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - (a + b))2 + (y - (b - a))2 = (x - (a - b))2 + (y - (a + b))2
⇒ x2 - 2(a + b)x + (a + b)2 + y2 - 2(b - a)y + (b - a)2 = x2 - 2(a - b)x + (a - b)2 + y2 - 2(a + b)y + (a + b)2
⇒ x(- 2a - 2b + 2a - 2b) = y(2b - 2a - 2a - 2b)
⇒ x(- 4b) = y(- 4a)
⇒ x(b) = y(a)
⇒
Applying componendo and dividendo,
⇒ $\frac{x-y}{x+y}=\frac{a-b}{a+b}$
∴ Thus proved.
Given points are A(3, 4), B(8, - 6) and C(13, 9).
Let us find the distance between sides AB, BC and CA.
We know that distance between the two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
⇒ AB = √125
⇒ BC = √250
⇒ CA = √125
Now,
$\Rightarrow \mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{125})^{2}+(\sqrt{125})^{2}$
⇒ AB2 + CA2 = 125 + 125
⇒ AB2 + CA2 = 250
$\Rightarrow \mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{250})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right angled isosceles triangle.
Given points are A(1, 1), B(- √3, √3) and C(- 1, - 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the two points (x1, y1) and (x2, y2) is
Now,
⇒ SA = SB
⇒ SA2 = SB2
⇒ (x - (a + b))2 + (y - (b - a))2 = (x - (a - b))2 + (y - (a + b))2
⇒ x2 - 2(a + b)x + (a + b)2 + y2 - 2(b - a)y + (b - a)2 = x2 - 2(a - b)x + (a - b)2 + y2 - 2(a + b)y + (a + b)2
⇒ x(- 2a - 2b + 2a - 2b) = y(2b - 2a - 2a - 2b)
⇒ x(- 4b) = y(- 4a)
⇒ x(b) = y(a)
⇒
Applying componendo and dividendo,
⇒ $\frac{x-y}{x+y}=\frac{a-b}{a+b}$
∴ Thus proved.
Question 13
Prove that the points (3, 4), (8, - 6) and (13, 9) are the vertices of a right angled triangle.
Sol :Given points are A(3, 4), B(8, - 6) and C(13, 9).
Let us find the distance between sides AB, BC and CA.
We know that distance between the two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
$\Rightarrow A B=\sqrt{(3-8)^{2}+(4-(-6))^{2}}$
$\Rightarrow A B=\sqrt{(3-8)^{2}+(4+6)^{2}}$
$\Rightarrow A B=\sqrt{(-5)^{2}+(10)^{2}}$
$\Rightarrow A B=\sqrt{25+100}$
⇒ AB = √125
$\Rightarrow \mathrm{BC}=\sqrt{(8-13)^{2}+(-6-9)^{2}}$
$\Rightarrow \mathrm{BC}=\sqrt{(-5)^{2}+(-15)^{2}}$
$\Rightarrow \mathrm{BC}=\sqrt{25+225}$
⇒ BC = √250
$\Rightarrow C A=\sqrt{(13-3)^{2}+(9-4)^{2}}$
$\Rightarrow C A=\sqrt{(10)^{2}+(5)^{2}}$
$\Rightarrow C A=\sqrt{100+25}$
⇒ CA = √125
Now,
$\Rightarrow \mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{125})^{2}+(\sqrt{125})^{2}$
⇒ AB2 + CA2 = 125 + 125
⇒ AB2 + CA2 = 250
$\Rightarrow \mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{250})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right angled isosceles triangle.
Question 14 A
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following
triangles whose vertices are:
(1, 1), (-
), (- 1, -
1)
Sol :(1, 1), (-
), (- 1, -
1)
Given points are A(1, 1), B(- √3, √3) and C(- 1, - 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ $A B=\sqrt{(1-(-\sqrt{3}))^{2}+(1-\sqrt{3})^{2}}$
⇒ $A B=\sqrt{(1+\sqrt{3})^{2}+(1-\sqrt{3})^{2}}$
⇒ $A B=\sqrt{1+2 \sqrt{3}+3+1-2 \sqrt{3}+3}$
⇒ $A B=\sqrt{8}$
⇒ $\mathrm{BC}=\sqrt{(-\sqrt{3}-(-1))^{2}+(\sqrt{3}-(-1))^{2}}$
⇒ $B C=\sqrt{(1-\sqrt{3})^{2}+(1+\sqrt{3})^{2}}$
⇒ $\mathrm{BC}=\sqrt{1-2 \sqrt{3}+3+1+2 \sqrt{3}+3}$
⇒ $\mathrm{BC}=\sqrt{8}$
⇒ $\mathrm{CA}=\sqrt{(-1-1)^{2}+(-1-1)^{2}}$
We got AB = BC = CA
∴ The given points form an equilateral triangle.
Given points are A(0, 2), B(7, 0) and C(2, 5).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒ $A B=\sqrt{(1-(-\sqrt{3}))^{2}+(1-\sqrt{3})^{2}}$
⇒ $A B=\sqrt{(1+\sqrt{3})^{2}+(1-\sqrt{3})^{2}}$
⇒ $A B=\sqrt{1+2 \sqrt{3}+3+1-2 \sqrt{3}+3}$
⇒ $A B=\sqrt{8}$
⇒ $\mathrm{BC}=\sqrt{(-\sqrt{3}-(-1))^{2}+(\sqrt{3}-(-1))^{2}}$
⇒ $B C=\sqrt{(1-\sqrt{3})^{2}+(1+\sqrt{3})^{2}}$
⇒ $\mathrm{BC}=\sqrt{1-2 \sqrt{3}+3+1+2 \sqrt{3}+3}$
⇒ $\mathrm{BC}=\sqrt{8}$
⇒ $\mathrm{CA}=\sqrt{(-1-1)^{2}+(-1-1)^{2}}$
⇒$\mathrm{CA}=\sqrt{(-2)^{2}+(-2)^{2}}$
⇒$\mathrm{CA}=\sqrt{4+4}$
⇒$\mathrm{CA}=\sqrt{8}$
We got AB = BC = CA
∴ The given points form an equilateral triangle.
Question 14 B
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following
triangles whose vertices are:
(0, 2), (7, 0), (2, 5)
Sol :Given points are A(0, 2), B(7, 0) and C(2, 5).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ $\mathrm{AB}=\sqrt{(0-7)^{2}+(2-0)^{2}}$
We got AB≠BC≠CA
∴ The given points form a scalene triangle.
Given points are A(- 2, 5), B(7, 10) and C(3, - 4).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒ $\mathrm{AB}=\sqrt{(0-7)^{2}+(2-0)^{2}}$
⇒$\mathrm{AB}=\sqrt{(-7)^{2}+(2)^{2}}$
⇒$\mathrm{AB}=\sqrt{49+4}$
⇒$\mathrm{AB}=\sqrt{53}$
⇒$\mathrm{BC}=\sqrt{(7-2)^{2}+(0-5)^{2}}$
⇒ $B C=\sqrt{(5)^{2}+(-5)^{2}}$⇒$B C=\sqrt{25+25}$
⇒$B C=\sqrt{50}$
⇒$C A=\sqrt{(2-0)^{2}+(5-2)^{2}}$
⇒$C A=\sqrt{(2)^{2}+(3)^{2}}$
⇒$C A=\sqrt{4+9}$
⇒$C A=\sqrt{13}$
We got AB≠BC≠CA
∴ The given points form a scalene triangle.
Question 14 C
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following
triangles whose vertices are:
(- 2, 5), (7, 10), (3, - 4)
Sol :Given points are A(- 2, 5), B(7, 10) and C(3, - 4).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
⇒$\mathrm{AB}=\sqrt{(-2-7)^{2}+(5-10)^{2}}$
We got AB = CA
Now,
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{106})^{2}+(\sqrt{106})^{2}$
⇒ AB2 + CA2 = 106 + 106
⇒ AB2 + CA2 = 212
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{212})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right angles isosceles triangle.
Sol :
Given points are A(4, 4), B(3, 5) and C(- 1, - 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒$\mathrm{AB}=\sqrt{(-2-7)^{2}+(5-10)^{2}}$
⇒$\mathrm{AB}=\sqrt{(-9)^{2}+(-5)^{2}}$
⇒$\mathrm{AB}=\sqrt{81+25}$
⇒$\mathrm{AB}=\sqrt{106}$
⇒$\mathrm{BC}=\sqrt{(7-3)^{2}+(10-(-4))^{2}}$
⇒$\mathrm{BC}=\sqrt{(7-3)^{2}+(10+4)^{2}}$
⇒$\mathrm{BC}=\sqrt{4^{2}+14^{2}}$
⇒$\mathrm{BC}=\sqrt{16+196}$
⇒$B C=\sqrt{212}$⇒$C A=\sqrt{(3-(-2))^{2}+(-4-5)^{2}}$
⇒$C A=\sqrt{(3+2)^{2}+(-4-5)^{2}}$
⇒$C A=\sqrt{(5)^{2}+(-9)^{2}}$
⇒$C A=\sqrt{25+81}$
⇒$C A=\sqrt{106}$
We got AB = CA
Now,
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{106})^{2}+(\sqrt{106})^{2}$
⇒ AB2 + CA2 = 106 + 106
⇒ AB2 + CA2 = 212
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{212})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right angles isosceles triangle.
Question 14 D
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following
triangles whose vertices are:
(4, 4), (3, 5), (- 1, - 1)Sol :
Given points are A(4, 4), B(3, 5) and C(- 1, - 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒ $A B=\sqrt{(4-3)^{2}+(4-5)^{2}}$
Now,
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{2})^{2}+(\sqrt{50})^{2}$
⇒ AB2 + CA2 = 2 + 50
⇒ AB2 + CA2 = 52
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{52})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right - angled triangle.
), (3, 0), (- 1, 0)
Sol :
Given points are A(1, 2
), B(3, 0) and
C(- 1, 0).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒ $A B=\sqrt{(4-3)^{2}+(4-5)^{2}}$
⇒$A B=\sqrt{(1)^{2}+(-1)^{2}}$
⇒$A B=\sqrt{1+1}$
⇒$A B=\sqrt{2}$
⇒$B C=\sqrt{(3-(-1))^{2}+(5-(-1))^{2}}$
⇒$B C=\sqrt{(3+1)^{2}+(5+1)^{2}}$
⇒$B C=\sqrt{(4)^{2}+(6)^{2}}$
⇒$B C=\sqrt{16+36}$
⇒$B C=\sqrt{52}$⇒$C A=\sqrt{(-1-4)^{2}+(-1-4)^{2}}$
⇒$C A=\sqrt{(-5)^{2}+(-5)^{2}}$
⇒$C A=\sqrt{25+25}$
⇒$C A=\sqrt{50}$
Now,
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{2})^{2}+(\sqrt{50})^{2}$
⇒ AB2 + CA2 = 2 + 50
⇒ AB2 + CA2 = 52
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{52})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right - angled triangle.
Question 14 E
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following
triangles whose vertices are:
(1, 2), (3, 0), (- 1, 0)Sol :
Given points are A(1, 2
), B(3, 0) and
C(- 1, 0).Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒$\mathrm{AB}=\sqrt{(1-3)^{2}+(2 \sqrt{3}-0)^{2}}$
⇒ AB = 4
⇒ $\mathrm{BC}=\sqrt{(3-(-1))^{2}+(0-0)^{2}}$
⇒ CA = 4
We got AB = BC = CA
∴ The given points form an equilateral triangle.
Sol :
Given points are A(0, 6), B(- 5, 3) and C(3, 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒$\mathrm{AB}=\sqrt{(1-3)^{2}+(2 \sqrt{3}-0)^{2}}$
⇒$\mathrm{AB}=\sqrt{(-2)^{2}+(2 \sqrt{3})^{2}}$
⇒$\mathrm{AB}=\sqrt{4+12}$
⇒$\mathrm{AB}=\sqrt{16}$
⇒ AB = 4
⇒ $\mathrm{BC}=\sqrt{(3-(-1))^{2}+(0-0)^{2}}$
⇒$\mathrm{BC}=\sqrt{(3+1)^{2}+(0-0)^{2}}$
⇒$\mathrm{BC}=\sqrt{4^{2}}$
⇒$\mathrm{BC}=4$
⇒$\mathrm{CA}=\sqrt{(-1-1)^{2}+(0-2 \sqrt{3})^{2}}$
⇒$\mathrm{CA}=\sqrt{(-2)^{2}+(-2 \sqrt{3})^{2}}$
⇒$\mathrm{CA}=\sqrt{4+12}$
⇒$\mathrm{CA}=\sqrt{16}$
⇒ CA = 4
We got AB = BC = CA
∴ The given points form an equilateral triangle.
Question 14 F
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following
triangles whose vertices are:
(0, 6), (- 5, 3), (3, 1)Sol :
Given points are A(0, 6), B(- 5, 3) and C(3, 1).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
⇒$A B=\sqrt{(0-(-5))^{2}+(6-3)^{2}}$
We got AB = CA
Now,
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{34})^{2}+(\sqrt{34})^{2}$
⇒ AB2 + CA2 =34+34
⇒ AB2 + CA2 = 68
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{68})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right - angled isosceles triangle.
Sol :
Given points are A(5, - 2), B(6, 4) and C(7, - 2).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
⇒$A B=\sqrt{(0-(-5))^{2}+(6-3)^{2}}$
⇒$A B=\sqrt{(0+5)^{2}+(6-3)^{2}}$
⇒$A B=\sqrt{5^{2}+3^{2}}$
⇒$A B=\sqrt{25+9}$
⇒$A B=\sqrt{34}$
⇒$B C=\sqrt{(-5-3)^{2}+(3-1)^{2}}$
⇒$B C=\sqrt{(-8)^{2}+(2)^{2}}$
⇒$A B=\sqrt{64+4}$
⇒$\mathrm{BC}=\sqrt{68}$⇒$\mathrm{CA}=\sqrt{(3-0)^{2}+(1-6)^{2}}$
⇒$\mathrm{CA}=\sqrt{(3)^{2}+(-5)^{2}}$
⇒$\mathrm{CA}=\sqrt{9+25}$
⇒$\mathrm{CA}=\sqrt{34}$
We got AB = CA
Now,
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{34})^{2}+(\sqrt{34})^{2}$
⇒ AB2 + CA2 =34+34
⇒ AB2 + CA2 = 68
⇒ $\mathrm{AB}^{2}+\mathrm{CA}^{2}=(\sqrt{68})^{2}$
⇒ AB2 + CA2 = BC2
∴ The given points form a right - angled isosceles triangle.
Question 14 G
Determine the type (isosceles, right angled, right angled isosceles, equilateral, scalene) of the following
triangles whose vertices are:
(5, - 2), (6, 4), (7, - 2)Sol :
Given points are A(5, - 2), B(6, 4) and C(7, - 2).
Let us find the distance between sides AB, BC and CA.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
⇒$A B=\sqrt{(5-6)^{2}+(-2-4)^{2}}$
⇒$\mathrm{BC}=\sqrt{1+36}$
We got AB = BC≠CA
∴ The given points form an isosceles triangle.
Given points are A(at2, 2at), B$\left(\frac{a}{t^{2}}, \frac{2 a}{t}\right)$ and C(a, 0).
Let us find the distance between sides AB, BC and CA.
We know that distance between the two points (x1, y1) and (x2, y2) is
⇒$A B=\sqrt{(5-6)^{2}+(-2-4)^{2}}$
⇒$A B=\sqrt{(-1)^{2}+(-6)^{2}}$
⇒$A B=\sqrt{1+36}$
⇒$A B=\sqrt{37}$
⇒$B C=\sqrt{(6-7)^{2}+(4-(-2))^{2}}$
⇒$B C=\sqrt{(6-7)^{2}+(4+2)^{2}}$
⇒$B C=\sqrt{(-1)^{2}+(6)^{2}}$
⇒$\mathrm{BC}=\sqrt{1+36}$
⇒$\mathrm{BC}=\sqrt{37}$
⇒$\mathrm{CA}=\sqrt{(7-5)^{2}+(-2-(-2))^{2}}$
⇒$\mathrm{CA}=\sqrt{(7-5)^{2}+(-2+2)^{2}}$
⇒$\mathrm{CA}=\sqrt{(2)^{2}+(0)^{2}}$
⇒$\mathrm{CA}=\sqrt{4}$
⇒ CA=2We got AB = BC≠CA
∴ The given points form an isosceles triangle.
Question 15
If A(at2, 2at), B$\left(\frac{a}{t^{2}}, \frac{2
a}{t}\right)$ and C(a, 0) be any three points, show that $\frac{1}{A C}+\frac{1}{B C}$ is independent of
t.
Sol :Given points are A(at2, 2at), B$\left(\frac{a}{t^{2}}, \frac{2 a}{t}\right)$ and C(a, 0).
Let us find the distance between sides AB, BC and CA.
We know that distance between the two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
⇒ AC = at2 + a
$\Rightarrow A C=\sqrt{\left(a t^{2}-a\right)^{2}+(2 a t-0)^{2}}$
$\Rightarrow A C=\sqrt{a^{2} t^{4}+a^{2}-2 a^{2} t^{2}+4 a^{2} t^{2}}$
$\Rightarrow A C=\sqrt{a^{2} t^{4}+2 a^{2} t^{2}+a^{2}}$
$\Rightarrow A C=\sqrt{\left(a t^{2}+a\right)^{2}}$
⇒ AC = at2 + a
$\Rightarrow B C=\sqrt{\left(\frac{a}{t^{2}}-a\right)^{2}+\left(\frac{2
a}{t}-0\right)^{2}}$
$\Rightarrow B C=\sqrt{\frac{a^{2}}{t^{4}}+a^{2}-\frac{2 a^{2}}{t^{2}}+\frac{4
a^{2}}{t^{2}}}$
$\Rightarrow B C=\sqrt{\frac{a^{2}}{t^{4}}+\frac{2 a^{2}}{t^{2}}+a^{2}}$
$\Rightarrow B C=\sqrt{\left(\frac{a}{t^{2}}+a\right)^{2}}$
$\Rightarrow B C=\frac{a}{t^{2}}+a$
$\Rightarrow \mathrm{BC}=\frac{\mathrm{a}+\mathrm{at}^{2}}{\mathrm{t}^{2}}$
Now,
$\Rightarrow
\frac{1}{\mathrm{AC}}+\frac{1}{\mathrm{BC}}=\frac{1}{\mathrm{at}^{2}+\mathrm{a}}+\frac{1}{\frac{\mathrm{a}+\mathrm{at}^{2}}{\mathrm{t}^{2}}}$
$\Rightarrow
\frac{1}{\mathrm{AC}}+\frac{1}{\mathrm{BC}}=\frac{1}{\mathrm{at}^{2}+\mathrm{a}}+\frac{\mathrm{t}^{2}}{\mathrm{a}+\mathrm{at}^{2}}$
$\Rightarrow
\frac{1}{\mathrm{AC}}+\frac{1}{\mathrm{BC}}=\frac{1+\mathrm{t}^{2}}{\mathrm{a}\left(1+\mathrm{t}^{2}\right)}$
$\Rightarrow \frac{1}{\mathrm{AC}}+\frac{1}{\mathrm{BC}}=\frac{1}{\mathrm{a}}$
$\therefore \frac{1}{A C}+\frac{1}{B C}$ is independent of t.
Question 16
If two vertices of an equilateral triangle be (0, 0) and (3, √3), find the co - ordinates of the third
vertex.
Sol :
Given that A(0, 0) and B(3, √3) are two vertices of an equilateral triangle.
Let us assume C(x, y) be the third vertex of the triangle.
We have AB = BC = CA
We know that the distance between the two points (x1, y1) and (x2, y2) is
Let us assume C(x, y) be the third vertex of the triangle.
We have AB = BC = CA
We know that the distance between the two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒ BC = CA
⇒ BC2 = CA2
⇒ (3 - x)2 + (√3 - y)2 = (x - 0)2 + (y - 0)2
⇒ x2 - 6x + 9 + 3 + y2 - 2√3y = x2 + y2
⇒ 6x = 12 - 2√3y
⇒ $\mathrm{X}=\frac{12-2 \sqrt{3} \mathrm{y}}{6}$ ..... - (1)
⇒ AB = BC
⇒ AB2 = BC2
⇒ (0 - 3)2 + (0 - √3)2 = (3 - x)2 + (√3 - y)2
⇒ 9 + 3 = 9 - 6x + x2 + 3 - 2√3y + y2
From (1)
$\Rightarrow \mathrm{y}^{2}-2 \sqrt{3} \mathrm{y}+\left(\frac{12-2 \sqrt{3} \mathrm{y}}{6}\right)^{2}-6\left(\frac{12-2 \sqrt{3} \mathrm{y}}{6}\right)=0$
$\Rightarrow \mathrm{y}^{2}-2 \sqrt{3} \mathrm{y}+\left(\frac{144+12 \mathrm{y}^{2}-48 \sqrt{3} \mathrm{y}}{36}\right)-12+2 \sqrt{3} \mathrm{y}=0$
⇒ $\Rightarrow 36 y^{2}-432+144+12 y^{2}-48 \sqrt{3} y=0$
⇒ 48y2 - 48√3y - 288 = 0
⇒ y2–√3y –6 = 0
⇒ y2 - 2√3y + √3y - 6 = 0
⇒ y(y - 2√3) + √3(y - 2√3) = 0
⇒ (y + √3)(y - 2√3) = 0
⇒ y + √3 = 0 (or) y - 2√3 = 0
⇒ y = - √3 (or) y = 2√3
From (1), for y = √3
From (1), for y = 2√3
∴ The third vertex of equilateral triangle is (0, 2√3) and (3, √3).
Now,
⇒ BC = CA
⇒ BC2 = CA2
⇒ (3 - x)2 + (√3 - y)2 = (x - 0)2 + (y - 0)2
⇒ x2 - 6x + 9 + 3 + y2 - 2√3y = x2 + y2
⇒ 6x = 12 - 2√3y
⇒ $\mathrm{X}=\frac{12-2 \sqrt{3} \mathrm{y}}{6}$ ..... - (1)
⇒ AB = BC
⇒ AB2 = BC2
⇒ (0 - 3)2 + (0 - √3)2 = (3 - x)2 + (√3 - y)2
⇒ 9 + 3 = 9 - 6x + x2 + 3 - 2√3y + y2
From (1)
$\Rightarrow \mathrm{y}^{2}-2 \sqrt{3} \mathrm{y}+\left(\frac{12-2 \sqrt{3} \mathrm{y}}{6}\right)^{2}-6\left(\frac{12-2 \sqrt{3} \mathrm{y}}{6}\right)=0$
$\Rightarrow \mathrm{y}^{2}-2 \sqrt{3} \mathrm{y}+\left(\frac{144+12 \mathrm{y}^{2}-48 \sqrt{3} \mathrm{y}}{36}\right)-12+2 \sqrt{3} \mathrm{y}=0$
⇒ $\Rightarrow 36 y^{2}-432+144+12 y^{2}-48 \sqrt{3} y=0$
⇒ 48y2 - 48√3y - 288 = 0
⇒ y2–√3y –6 = 0
⇒ y2 - 2√3y + √3y - 6 = 0
⇒ y(y - 2√3) + √3(y - 2√3) = 0
⇒ (y + √3)(y - 2√3) = 0
⇒ y + √3 = 0 (or) y - 2√3 = 0
⇒ y = - √3 (or) y = 2√3
From (1), for y = √3
$\Rightarrow x=\frac{12-2 \sqrt{3}(-\sqrt{3})}{6}$
$\Rightarrow x=\frac{12+6}{6}$
$\Rightarrow x=\frac{18}{6}$
⇒ x = 3From (1), for y = 2√3
$\Rightarrow x=\frac{12-2 \sqrt{3}(2 \sqrt{3})}{6}$
$\Rightarrow x=\frac{12-12}{6}$
⇒ x = 0∴ The third vertex of equilateral triangle is (0, 2√3) and (3, √3).
Question 17 A
Find the circum - centre and circum - radius of the triangle whose vertices are
(- 2, 3), (2, - 1) and (4, 0).
Sol :Given that we need to find the circum - centre and circum - radius of the triangle whose vertices are A(- 2, 3), B(2, - 1), C(4, 0).
Let us assume O(x, y) be the Circum - centre of the circle.
We know that distance from circum - centre to any vertex is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - (- 2))2 + (y - 3)2 = (x - 2)2 + (y - (- 1))2
⇒ (x + 2)2 + (y - 3)2 = (x - 2)2 + (y + 1)2
⇒ x2 + 4x + 4 + y2 - 6y + 9 = x2 - 4x + 4 + y2 + 2y + 1
⇒ 8x - 8y = - 8
⇒ x - y = - 1 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 2)2 + (y - (- 1))2 = (x - 4)2 + (y - 0)2
⇒ (x - 2)2 + (y + 1)2 = (x - 4)2 + (y)2
⇒ x2 - 4x + 4 + y2 + 2y + 1 = x2 - 8x + 16 + y2
⇒ 4x + 2y = 11 .... - (2)
On solving (1) and (2), we get
⇒ $\mathrm{x}=\frac{3}{2} \mathrm{and} \mathrm{y}=\frac{5}{2}$
∴ $\left(\frac{3}{2}, \frac{5}{2}\right)$ is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the circum - radius of the circle.
∴ The radius of the circle is $\frac{5 \sqrt{2}}{2}$
Given that circle passes through the points A(6, - 6), B(3, - 7), C(3, 3).
Let us assume O(x, y) be the centre of the circle.
We know that distance from the centre to any point on h circle is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - (- 2))2 + (y - 3)2 = (x - 2)2 + (y - (- 1))2
⇒ (x + 2)2 + (y - 3)2 = (x - 2)2 + (y + 1)2
⇒ x2 + 4x + 4 + y2 - 6y + 9 = x2 - 4x + 4 + y2 + 2y + 1
⇒ 8x - 8y = - 8
⇒ x - y = - 1 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 2)2 + (y - (- 1))2 = (x - 4)2 + (y - 0)2
⇒ (x - 2)2 + (y + 1)2 = (x - 4)2 + (y)2
⇒ x2 - 4x + 4 + y2 + 2y + 1 = x2 - 8x + 16 + y2
⇒ 4x + 2y = 11 .... - (2)
On solving (1) and (2), we get
⇒ $\mathrm{x}=\frac{3}{2} \mathrm{and} \mathrm{y}=\frac{5}{2}$
∴ $\left(\frac{3}{2}, \frac{5}{2}\right)$ is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the circum - radius of the circle.
$\Rightarrow
\mathrm{r}=\mathrm{OA}=\sqrt{\left(\frac{3}{2}-(-2)\right)^{2}+\left(\frac{5}{2}-3\right)^{2}}$
$\Rightarrow \mathrm{r}=\sqrt{\left(\frac{7}{2}\right)^{2}+\left(-\frac{1}{2}\right)^{2}}$
$\Rightarrow \mathrm{r}=\sqrt{\frac{49}{4}+\frac{1}{4}}$
$\Rightarrow \mathrm{r}=\sqrt{\frac{50}{4}}$
$\Rightarrow \mathrm{r}=\sqrt{\frac{25 \times 2}{4}}$
$\Rightarrow \mathrm{r}=\frac{5 \sqrt{2}}{2}$
∴ The radius of the circle is $\frac{5 \sqrt{2}}{2}$
Question 17 B
Find the centre of a circle passing through the points
(6, - 6), (3, - 7) and (3, 3).
Sol :Given that circle passes through the points A(6, - 6), B(3, - 7), C(3, 3).
Let us assume O(x, y) be the centre of the circle.
We know that distance from the centre to any point on h circle is equal.
So, OA = OB = OC
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - 6)2 + (y - (- 6))2 = (x - 3)2 + (y - (- 7))2
⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2
⇒ x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49
⇒ 6x + 2y = 14
⇒ 3x + y = 7 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 3)2 + (y - (- 7))2 = (x - 3)2 + (y - 3)2
⇒ (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2
⇒ x2 - 6x + 9 + y2 + 14y + 49 = x2 - 6x + 9 + y2 - 6y + 9
⇒ 20y = - 40
⇒ y = - 2 .... - (2)
Substituting (2) in (1), we get
⇒ x = 3
∴ (3, - 2) is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the radius of the circle.
⇒ r = √(9 + 16)
⇒ r = √25
⇒ r = 5
∴ The radius of the circle is 5.
Given that the circle has centre O(2x - 1, 3x + 1) and passes through the point A(- 3, - 1) and has a radius(r) of 10 units.
We know that the radius of the circle is the distance between the centre and any point on the circle.
So, we have r = OA
⇒ OA = 10
⇒ OA2 = 100
⇒ (2x - 1 - (- 3))2 + (3x + 1 - (- 1))2 = 100
⇒ (2x + 2)2 + (3x + 2)2 = 100
⇒ 4x2 + 8x + 4 + 9x2 + 12x + 4 = 100
⇒ 13x2 + 20x - 92 = 0
⇒ 13x2 - 26x + 46x - 92 = 0
⇒ 13x(x - 2) + 46(x - 2) = 0
⇒ (13x + 46)(x - 2) = 0
⇒ 13x + 46 = 0 (or) x - 2 = 0
⇒ 13x = - 46 (or) x = 2
$\Rightarrow x=\frac{-46}{13}($ or $) x=2$
∴ The values of the x are $\frac{-46}{13}$ or 2.
Given points are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).
We need to prove that these are the vertices of a square.
We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
Now,
⇒ OA = OB
⇒ OA2 = OB2
⇒ (x - 6)2 + (y - (- 6))2 = (x - 3)2 + (y - (- 7))2
⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2
⇒ x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49
⇒ 6x + 2y = 14
⇒ 3x + y = 7 ..... (1)
Now,
⇒ OB = OC
⇒ OB2 = OC2
⇒ (x - 3)2 + (y - (- 7))2 = (x - 3)2 + (y - 3)2
⇒ (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2
⇒ x2 - 6x + 9 + y2 + 14y + 49 = x2 - 6x + 9 + y2 - 6y + 9
⇒ 20y = - 40
⇒ y = - 2 .... - (2)
Substituting (2) in (1), we get
⇒ x = 3
∴ (3, - 2) is the centre of the circle.
We know radius is the distance between the centre and any point on the circle.
Let ‘r’ be the radius of the circle.
$\Rightarrow \mathrm{r}=\mathrm{OA}=\sqrt{(3-6)^{2}+(-2-(-6))^{2}}$
$\Rightarrow \mathrm{r}=\sqrt{(-3)^{2}+(4)^{2}}$
⇒ r = √(9 + 16)
⇒ r = √25
⇒ r = 5
∴ The radius of the circle is 5.
Question 18
If the line segment joining the points A(a, b) and B(c, d) subtends a right angle at the origin, show that
ac + bd = 0.
Sol :
Given that the line segment joining the points A(a, b) and B(c, d) subtends a right angle at the origin O(0,
0)
So, AOB is a right angled triangle with right angle at O.
We got OA2 + OB2 = AB2 [By Pythagoras Theorem]
We know that the distance between the points (x1, y1) and (x2, y2) is
So, AOB is a right angled triangle with right angle at O.
We got OA2 + OB2 = AB2 [By Pythagoras Theorem]
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
.
⇒ OA2 + OB2 = AB2
⇒ (0 - a)2 + (0 - b)2 + (0 - c)2 + (0 - d)2 = (a - c)2 + (b - d)2
⇒ a2 + b2 + c2 + d2 = a2 + c2 - 2ac + b2 + d2 - 2bd
⇒ 2ac + 2bd = 0
⇒ ac + bd = 0
⇒ OA2 + OB2 = AB2
⇒ (0 - a)2 + (0 - b)2 + (0 - c)2 + (0 - d)2 = (a - c)2 + (b - d)2
⇒ a2 + b2 + c2 + d2 = a2 + c2 - 2ac + b2 + d2 - 2bd
⇒ 2ac + 2bd = 0
⇒ ac + bd = 0
Question 19
The centre of the circle is (2x - 1, 3x + 1) and radius is 10 units. Find the value of x if the circle
passes through the point (- 3, - 1).
Sol :Given that the circle has centre O(2x - 1, 3x + 1) and passes through the point A(- 3, - 1) and has a radius(r) of 10 units.
We know that the radius of the circle is the distance between the centre and any point on the circle.
So, we have r = OA
⇒ OA = 10
⇒ OA2 = 100
⇒ (2x - 1 - (- 3))2 + (3x + 1 - (- 1))2 = 100
⇒ (2x + 2)2 + (3x + 2)2 = 100
⇒ 4x2 + 8x + 4 + 9x2 + 12x + 4 = 100
⇒ 13x2 + 20x - 92 = 0
⇒ 13x2 - 26x + 46x - 92 = 0
⇒ 13x(x - 2) + 46(x - 2) = 0
⇒ (13x + 46)(x - 2) = 0
⇒ 13x + 46 = 0 (or) x - 2 = 0
⇒ 13x = - 46 (or) x = 2
$\Rightarrow x=\frac{-46}{13}($ or $) x=2$
∴ The values of the x are $\frac{-46}{13}$ or 2.
Question 20 A
Prove that the points (4, 3), (6, 4), (5, 6) and (3, 5) are the vertices of a square.
Sol :Given points are A(4, 3), B(6, 4), C(5, 6) and D(3, 5).
We need to prove that these are the vertices of a square.
We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒ AB = √5
⇒ BC = √5
⇒ CD = √5
⇒ DA = √5
We got AB = BC = CD = DA, this may be square (or) rhombus.
Now we find the lengths of the diagonals.
⇒ AC = √10
⇒ BD = √10
We got AC = BD.
∴ The points form a square.
Given points are A(4, 3), B(6, 4), C(5, 6) and D(- 4, 4).
We need to prove that these are the vertices of a square.
We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
Now,
$\Rightarrow A B=\sqrt{(4-6)^{2}+(3-4)^{2}}$
$\Rightarrow A B=\sqrt{(-2)^{2}+(-1)^{2}}$
$\Rightarrow A B=\sqrt{4+1}$
⇒ AB = √5
$\Rightarrow B C=\sqrt{(6-5)^{2}+(4-6)^{2}}$
$\Rightarrow B C=\sqrt{(1)^{2}+(-2)^{2}}$
$\Rightarrow B C=\sqrt{1+4}$
⇒ BC = √5
$\Rightarrow C D=\sqrt{(5-3)^{2}+(6-5)^{2}}$
$\Rightarrow C D=\sqrt{(2)^{2}+(1)^{2}}$
$\Rightarrow C D=\sqrt{4+1}$
$\Rightarrow \mathrm{DA}=\sqrt{(3-4)^{2}+(5-3)^{2}}$
$\Rightarrow \mathrm{DA}=\sqrt{(-1)^{2}+(2)^{2}}$
$\Rightarrow \mathrm{DA}=\sqrt{1+4}$
⇒ DA = √5
We got AB = BC = CD = DA, this may be square (or) rhombus.
Now we find the lengths of the diagonals.
$\Rightarrow A C=\sqrt{(4-5)^{2}+(3-6)^{2}}$
$\Rightarrow A C=\sqrt{(-1)^{2}+(-3)^{2}}$
$\Rightarrow A C=\sqrt{1+9}$
⇒ AC = √10
$\Rightarrow B D=\sqrt{(6-3)^{2}+(4-5)^{2}}$
$\Rightarrow B D=\sqrt{(3)^{2}+(-1)^{2}}$
$\Rightarrow B D=\sqrt{9+1}$
⇒ BD = √10
We got AC = BD.
∴ The points form a square.
Question 20 B
Prove that the points (4, 3), (6, 4), (5, 6) and (- 4, 4) are the vertices of a square.
Sol :Given points are A(4, 3), B(6, 4), C(5, 6) and D(- 4, 4).
We need to prove that these are the vertices of a square.
We know that in the lengths of all sides are equal and the lengths of the diagonals are equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒$A B=\sqrt{(4-6)^{2}+(3-4)^{2}}$
We got AB = BC≠CD≠DA,
∴ The points doesn’t form a square.
Sol :
Given points are A(3, 2), B(6, 3), C(7, 6) and D(4, 5).
We need to prove that these are the vertices of a parallelogram.
We know that in the lengths of opposite sides are equal in a parallelogram.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
Now,
⇒$A B=\sqrt{(4-6)^{2}+(3-4)^{2}}$
⇒$A B=\sqrt{(-2)^{2}+(-1)^{2}}$
⇒$A B=\sqrt{4+1}$
⇒$A B=\sqrt{5}$
⇒$\mathrm{BC}=\sqrt{(6-5)^{2}+(4-6)^{2}}$⇒$\mathrm{BC}=\sqrt{(1)^{2}+(-2)^{2}}$
⇒$\mathrm{BC}=\sqrt{1+4}$
⇒$\mathrm{BC}=\sqrt{5}$
⇒$\mathrm{CD}=\sqrt{(5-(-4))^{2}+(6-4)^{2}}$
⇒$\mathrm{CD}=\sqrt{(9)^{2}+(2)^{2}}$
⇒$\mathrm{CD}=\sqrt{81+4}$
⇒$\mathrm{CD}=\sqrt{85}$
⇒$D A=\sqrt{(-4-4)^{2}+(4-3)^{2}}$⇒$D A=\sqrt{(-8)^{2}+(1)^{2}}$
⇒$D A=\sqrt{64+1}$
⇒$D A=\sqrt{65}$
We got AB = BC≠CD≠DA,
∴ The points doesn’t form a square.
Question 21
Prove that the points (3, 2), (6, 3), (7, 6), (4, 5) are the vertices of a parallelogram. Is it a rectangle?Sol :
Given points are A(3, 2), B(6, 3), C(7, 6) and D(4, 5).
We need to prove that these are the vertices of a parallelogram.
We know that in the lengths of opposite sides are equal in a parallelogram.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Now,
We got AB = CD and BC = DA, these are the vertices of a parallelogram.
Now we find the lengths of the diagonals.
We got AC≠BD.
∴ The points doesn’t form a rectangle.
Given points are A(6, 8), B(3, 7), C(- 2, - 2) and D(1, - 1).
We need to prove that these are the vertices of a parallelogram.
We know that in the lengths of opposite sides are equal in a parallelogram and the lengths of diagonals are not equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
Now,
$\Rightarrow A B=\sqrt{(3-6)^{2}+(2-3)^{2}}$
$\Rightarrow A B=\sqrt{(-3)^{2}+(-1)^{2}}$
$\Rightarrow A B=\sqrt{9+1}$
⇒ AB = √10$\Rightarrow B C=\sqrt{(6-7)^{2}+(3-6)^{2}}$
$\Rightarrow B C=\sqrt{(-1)^{2}+(-3)^{2}}$
$\Rightarrow B C=\sqrt{1+9}$
⇒ BC = √10$\Rightarrow C D=\sqrt{(7-4)^{2}+(6-5)^{2}}$
$\Rightarrow C D=\sqrt{(3)^{2}+(1)^{2}}$
$\Rightarrow C D=\sqrt{9+1}$
⇒ CD = √10$\Rightarrow \mathrm{DA}=\sqrt{(4-3)^{2}+(5-2)^{2}}$
$\Rightarrow \mathrm{DA}=\sqrt{(1)^{2}+(3)^{2}}$
$\Rightarrow \mathrm{DA}=\sqrt{1+9}$
⇒ DA = √10We got AB = CD and BC = DA, these are the vertices of a parallelogram.
Now we find the lengths of the diagonals.
$\Rightarrow A C=\sqrt{(3-7)^{2}+(2-6)^{2}}$
$\Rightarrow A C=\sqrt{(-4)^{2}+(-4)^{2}}$
$\Rightarrow A C=\sqrt{16+16}$
⇒ AC = √32$\Rightarrow B D=\sqrt{(6-4)^{2}+(3-5)^{2}}$
$\Rightarrow B D=\sqrt{(2)^{2}+(-2)^{2}}$
$\Rightarrow B D=\sqrt{4+4}$
⇒ BD = √8We got AC≠BD.
∴ The points doesn’t form a rectangle.
Question 22
Prove that the points (6, 8), (3, 7), (- 2, - 2), (1, - 1) are the vertices of a parallelogram.
Sol :Given points are A(6, 8), B(3, 7), C(- 2, - 2) and D(1, - 1).
We need to prove that these are the vertices of a parallelogram.
We know that in the lengths of opposite sides are equal in a parallelogram and the lengths of diagonals are not equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒$A B=\sqrt{(6-3)^{2}+(8-7)^{2}}$
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(6-(-2))^{2}+(8-(-2))^{2}}$
⇒$\mathrm{BD}=\sqrt{68}$
We got AC≠BD.
∴ The points form a parallelogram.
Given points are A(4, 8), B(0, 2), C(3, 0) and D(7, 6).
We need to prove that these are the vertices of a rectangle.
We know that in the lengths of opposite sides and lengths of diagonals are equal in a rectangle.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
Now,
⇒$A B=\sqrt{(6-3)^{2}+(8-7)^{2}}$
⇒$A B=\sqrt{(3)^{2}+(1)^{2}}$
⇒$A B=\sqrt{9+1}$
⇒$A B=\sqrt{10}$
⇒$\mathrm{BC}=\sqrt{(3-(-2))^{2}+(7-(-2))^{2}}$⇒$\mathrm{BC}=\sqrt{(5)^{2}+(9)^{2}}$
⇒$\mathrm{BC}=\sqrt{25+81}$
⇒$\mathrm{BC}=\sqrt{106}$
⇒$\mathrm{CD}=\sqrt{(-2-1)^{2}+(-2-(-1))^{2}}$
⇒$\mathrm{CD}=\sqrt{(-3)^{2}+(-1)^{2}}$
⇒$\mathrm{CD}=\sqrt{9+1}$
⇒$\mathrm{CD}=\sqrt{10}$
⇒$\mathrm{DA}=\sqrt{(1-6)^{2}+(-1-8)^{2}}$⇒$\mathrm{DA}=\sqrt{(-5)^{2}+(-9)^{2}}$
⇒$\mathrm{DA}=\sqrt{25+81}$
⇒$\mathrm{DA}=\sqrt{106}$
We got AB = CD and BC = DA, these are the vertices of a parallelogram or rectangle.Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(6-(-2))^{2}+(8-(-2))^{2}}$
⇒$A C=\sqrt{(8)^{2}+(10)^{2}}$
⇒$A C=\sqrt{64+100}$
⇒$A C=\sqrt{164}$
⇒$B D=\sqrt{(3-1)^{2}+(7-(-1))^{2}}$
⇒$B D=\sqrt{(2)^{2}+(8)^{2}}$
⇒$B D=\sqrt{4+64}$
⇒$\mathrm{BD}=\sqrt{68}$
We got AC≠BD.
∴ The points form a parallelogram.
Question 23
Prove that the points (4, 8), (0, 2), (3, 0) and (7, 6) are the vertices of a rectangle.
Sol :Given points are A(4, 8), B(0, 2), C(3, 0) and D(7, 6).
We need to prove that these are the vertices of a rectangle.
We know that in the lengths of opposite sides and lengths of diagonals are equal in a rectangle.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒ AB = √52
⇒ BC = √13
⇒ CD = √52
⇒ DA = √13
We got AB = CD and BC = DA, these are the vertices of a parallelogram or a rectangle.
Now we find the lengths of the diagonals.
We got AC = BD.
∴ The points form a rectangle.
⇒$A C=\sqrt{(1-2)^{2}+(0-7)^{2}}$
We got AC = BD.
∴ The points form a square not rhombus.
Sol :
Given points are A(4, 5), B(7, 6), C(4, 3) and D(1, 2).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
Now,
$\Rightarrow A B=\sqrt{(4-0)^{2}+(8-2)^{2}}$
$\Rightarrow A B=\sqrt{(4)^{2}+(6)^{2}}$
$\Rightarrow A B=\sqrt{16+36}$
⇒ AB = √52
$\Rightarrow B C=\sqrt{(0-3)^{2}+(2-0)^{2}}$
$\Rightarrow B C=\sqrt{(-3)^{2}+(2)^{2}}$
$\Rightarrow B C=\sqrt{9+4}$
⇒ BC = √13
$\Rightarrow C D=\sqrt{(3-7)^{2}+(0-6)^{2}}$
$\Rightarrow C D=\sqrt{(-4)^{2}+(-6)^{2}}$
$\Rightarrow C D=\sqrt{16+36}$
⇒ CD = √52
$\Rightarrow D A=\sqrt{(7-4)^{2}+(6-8)^{2}}$
$\Rightarrow D A=\sqrt{(3)^{2}+(-2)^{2}}$
$\Rightarrow D A=\sqrt{9+4}$
⇒ DA = √13
We got AB = CD and BC = DA, these are the vertices of a parallelogram or a rectangle.
Now we find the lengths of the diagonals.
$\Rightarrow A C=\sqrt{(4-3)^{2}+(8-0)^{2}}$
$\Rightarrow A C=\sqrt{(1)^{2}+(8)^{2}}$
$\Rightarrow A C=\sqrt{1+64}$
$\Rightarrow A C=\sqrt{65}$
$\Rightarrow B D=\sqrt{(0-7)^{2}+(2-6)^{2}}$
$\Rightarrow B D=\sqrt{(-7)^{2}+(-4)^{2}}$
$\Rightarrow B D=\sqrt{49+16}$
$\Rightarrow \mathrm{BD}=\sqrt{65}$We got AC = BD.
∴ The points form a rectangle.
Question 24
Show that the points A(1, 0), B(5, 3), C(2, 7) and D(- 2, 4) are the vertices of a rhombus.
Sol :
Given points are A(1, 0), B(5, 3), C(2, 7) and D(- 2, 4).
We need to prove that these are the vertices of a rhombus.
We know that in the lengths of sides are equal in a rhombus and the length of diagonals are not equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
We need to prove that these are the vertices of a rhombus.
We know that in the lengths of sides are equal in a rhombus and the length of diagonals are not equal.
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒$\mathrm{AB}=\sqrt{(1-5)^{2}+(0-3)^{2}}$
⇒ CD = 5
⇒$\mathrm{DA}=\sqrt{(-2-1)^{2}+(4-0)^{2}}$
Now,
⇒$\mathrm{AB}=\sqrt{(1-5)^{2}+(0-3)^{2}}$
⇒$\mathrm{AB}=\sqrt{(-4)^{2}+(-3)^{2}}$
⇒$\mathrm{AB}=\sqrt{16+9}$
⇒$\mathrm{AB}=\sqrt{25}$
⇒$\mathrm{AB}=5$
⇒$\mathrm{BC}=\sqrt{(5-2)^{2}+(3-7)^{2}}$
⇒$\mathrm{BC}=\sqrt{(3)^{2}+(4)^{2}}$
⇒$\mathrm{BC}=\sqrt{9+16}$
⇒$\mathrm{BC}=\sqrt{25}$
⇒$\mathrm{BC}=5$
⇒$\mathrm{CD}=\sqrt{(2-(-2))^{2}+(7-4)^{2}}$
⇒$\mathrm{CD}=\sqrt{(4)^{2}+(3)^{2}}$
⇒$\mathrm{CD}=\sqrt{16+9}$
⇒ $C D=\sqrt{25}$⇒ CD = 5
⇒$\mathrm{DA}=\sqrt{(-2-1)^{2}+(4-0)^{2}}$
⇒$\mathrm{DA}=\sqrt{(-3)^{2}+(4)^{2}}$
⇒$\mathrm{DA}=\sqrt{9+16}$
⇒$\mathrm{DA}=\sqrt{25}$
⇒ DA = 5
We got AB = BC = CD = DA, these are the vertices of a square or a rhombus.
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(1-2)^{2}+(0-7)^{2}}$
⇒$A C=\sqrt{(-1)^{2}+(-7)^{2}}$
⇒$A C=\sqrt{1+49}$
⇒$A C=\sqrt{50}$
⇒$B D=\sqrt{(5-(-2))^{2}+(3-4)^{2}}$
⇒$B D=\sqrt{(7)^{2}+(-1)^{2}}$
⇒$B D=\sqrt{49+1}$
⇒$B D=\sqrt{50}$
∴ The points form a square not rhombus.
Question 25 A
Name the type or quadrilateral formed, if any, by the following points and give reasons for your
answer:
(4, 5), (7, 6), (4, 3), (1, 2)Sol :
Given points are A(4, 5), B(7, 6), C(4, 3) and D(1, 2).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒$\mathrm{AB}=\sqrt{(4-7)^{2}+(5-6)^{2}}$
We got AB = CD and BC = DA, this may be a parallelogram or a rectangle.
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(4-4)^{2}+(5-3)^{2}}$
We got AC≠BD.
∴ The points form a parallelogram.
Sol :
Given points are A(- 1, - 2), B(1, 0), C(- 1, 2) and D(- 3, 0).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
Now,
⇒$\mathrm{AB}=\sqrt{(4-7)^{2}+(5-6)^{2}}$
⇒$\mathrm{AB}=\sqrt{(-3)^{2}+(-1)^{2}}$
⇒$\mathrm{AB}=\sqrt{9+1}$
⇒$\mathrm{AB}=\sqrt{10}$
⇒$\mathrm{BC}=\sqrt{(7-4)^{2}+(6-3)^{2}}$
⇒$\mathrm{BC}=\sqrt{(3)^{2}+(3)^{2}}$
⇒$\mathrm{BC}=\sqrt{9+9}$
⇒$\mathrm{BC}=\sqrt{18}$
⇒$\mathrm{CD}=\sqrt{(4-1)^{2}+(3-2)^{2}}$⇒$\mathrm{CD}=\sqrt{(3)^{2}+(1)^{2}}$
⇒$\mathrm{CD}=\sqrt{9+1}$
⇒$\mathrm{CD}=\sqrt{10}$
⇒$\mathrm{DA}=\sqrt{(1-4)^{2}+(2-5)^{2}}$
⇒$\mathrm{DA}=\sqrt{(-3)^{2}+(-3)^{2}}$
⇒$\mathrm{DA}=\sqrt{9+9}$
⇒$\mathrm{DA}=\sqrt{18}$
We got AB = CD and BC = DA, this may be a parallelogram or a rectangle.
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(4-4)^{2}+(5-3)^{2}}$
⇒$A C=\sqrt{(0)^{2}+(2)^{2}}$
⇒$A C=\sqrt{4}$
⇒$A C=2$
⇒$B D=\sqrt{(7-1)^{2}+(6-2)^{2}}$
⇒$B D=\sqrt{(6)^{2}+(4)^{2}}$
⇒$B D=\sqrt{52}$
We got AC≠BD.
∴ The points form a parallelogram.
Question 25 B
Name the type or quadrilateral formed, if any, by the following points and give reasons for your
answer:
(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)Sol :
Given points are A(- 1, - 2), B(1, 0), C(- 1, 2) and D(- 3, 0).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is
$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒$A B=\sqrt{(-1-1)^{2}+(-2-0)^{2}}$
⇒$\mathrm{BC}=\sqrt{8}$
We got AB = BC = CD = DA, this may be square (or) rhombus.
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(-1-(-1))^{2}+(-2-2)^{2}}$
We got AC = BD.
∴ The points form a square.
Sol :
Given points are A(- 3, 5), B(3, 1), C(0, 3) and D(- 1, - 4).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒$\mathrm{AB}=\sqrt{(-3-3)^{2}+(5-1)^{2}}$
We got AB≠BC≠CD≠DA, this may be a quadrilateral which is not of standard shape.
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(-3-0)^{2}+(5-3)^{2}}$
We got points ABC are collinear.
∴ The points doesn’t form a quadrilateral.
Given that A(- 1, 2) and C(3, 2) are the opposite vertices of a square.
Let us assume the other two vertices be B(x1, y1) and D(x2, y2) and the midpoint be M
We know that midpoint of AC = Midpoint of BD = M
$\Rightarrow \mathrm{M}=\left(\frac{-1+3}{2}, \frac{2+2}{2}\right)$
$\Rightarrow \mathrm{M}=\left(\frac{2}{2}, \frac{4}{2}\right)$
⇒ M = (1, 2)
$\Rightarrow\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)=(1,2)$
⇒ x1 + x2 = 2 ..... (1)
⇒ y1 + y2 = 4 ..... (2)
We know that lengths of the sides of the square are equal.
AB = BC = CD = DA.
We know that distance between two points (x1, y1) and (x2, y2) is
Now,
⇒$A B=\sqrt{(-1-1)^{2}+(-2-0)^{2}}$
⇒$A B=\sqrt{(-2)^{2}+(-2)^{2}}$
⇒$A B=\sqrt{4+4}$
⇒$A B=\sqrt{8}$
⇒$B C=\sqrt{(1-(-1))^{2}+(0-2)^{2}}$
⇒$B C=\sqrt{(2)^{2}+(-2)^{2}}$
⇒$B C=\sqrt{4+4}$
⇒$\mathrm{BC}=\sqrt{8}$
⇒$\mathrm{CD}=\sqrt{(-1-(-3))^{2}+(2-0)^{2}}$
⇒$\mathrm{CD}=\sqrt{(2)^{2}+(2)^{2}}$
⇒$\mathrm{CD}=\sqrt{4+4}$
⇒$\mathrm{DA}=\sqrt{(-3-(-1))^{2}+(0-(-2))^{2}}$
⇒$\mathrm{DA}=\sqrt{(-2)^{2}+(2)^{2}}$
⇒$\mathrm{DA}=\sqrt{4+4}$
⇒$\mathrm{DA}=\sqrt{8}$We got AB = BC = CD = DA, this may be square (or) rhombus.
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(-1-(-1))^{2}+(-2-2)^{2}}$
⇒$A C=\sqrt{(0)^{2}+(-4)^{2}}$
⇒$A C=\sqrt{16}$
⇒$A C=4$
⇒$B D=\sqrt{(1-(-3))^{2}+(0-0)^{2}}$
⇒$B D=\sqrt{(4)^{2}+(0)^{2}}$
⇒$B D=\sqrt{16}$
⇒$B D=4$
We got AC = BD.
∴ The points form a square.
Question 25 C
Name the type or quadrilateral formed, if any, by the following points and give reasons for your
answer:
(- 3, 5), (3, 1), (0, 3), (- 1, - 4)Sol :
Given points are A(- 3, 5), B(3, 1), C(0, 3) and D(- 1, - 4).
Let us find the lengths of the sides.
We know that the distance between the points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Now,
⇒$\mathrm{AB}=\sqrt{(-3-3)^{2}+(5-1)^{2}}$
⇒$\mathrm{AB}=\sqrt{(-6)^{2}+(4)^{2}}$
⇒$\mathrm{AB}=\sqrt{36+16}$
⇒$\mathrm{AB}=\sqrt{52}$
⇒$\mathrm{AB}=\sqrt{4 \times 13}$
⇒$\mathrm{BC}=2 \sqrt{13}$
⇒$\mathrm{BC}=\sqrt{(3)^{2}+(-2)^{2}}$
⇒$B C=\sqrt{9+4}$⇒$B C=\sqrt{13}$
⇒$C D=\sqrt{(0-(-1))^{2}+(3-(-4))^{2}}$
⇒CD $=\sqrt{(1)^{2}+(7)^{2}}$
⇒CD $=\sqrt{1+49}$
⇒CD $=\sqrt{50}$
⇒DA $=\sqrt{(-1-(-3))^{2}+(-4-5)^{2}}$
⇒$\mathrm{DA}=\sqrt{(2)^{2}+(-9)^{2}}$⇒$\mathrm{DA}=\sqrt{4+81}$
⇒$\mathrm{DA}=\sqrt{85}$
We got AB≠BC≠CD≠DA, this may be a quadrilateral which is not of standard shape.
Now we find the lengths of the diagonals.
⇒$A C=\sqrt{(-3-0)^{2}+(5-3)^{2}}$
⇒$A C=\sqrt{(-3)^{2}+(2)^{2}}$
⇒$A C=\sqrt{9+4}$
⇒$B C=\sqrt{13}$
⇒$B D=\sqrt{(4)^{2}+(5)^{2}}$
⇒$B D=\sqrt{(3-(-1))^{2}+(1-(-4))^{2}}$
⇒$B D=\sqrt{16+25}$
⇒$B D=\sqrt{41}$
⇒$\mathrm{AC}+\mathrm{BC}=\sqrt{13}+\sqrt{13}$
⇒$\mathrm{AC}+\mathrm{BC}=2 \sqrt{13}$
⇒ AC + BC = ABWe got points ABC are collinear.
∴ The points doesn’t form a quadrilateral.
Question 26
Two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two
vertices.
Sol :Given that A(- 1, 2) and C(3, 2) are the opposite vertices of a square.
Let us assume the other two vertices be B(x1, y1) and D(x2, y2) and the midpoint be M
We know that midpoint of AC = Midpoint of BD = M
$\Rightarrow \mathrm{M}=\left(\frac{-1+3}{2}, \frac{2+2}{2}\right)$
$\Rightarrow \mathrm{M}=\left(\frac{2}{2}, \frac{4}{2}\right)$
⇒ M = (1, 2)
$\Rightarrow\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)=(1,2)$
⇒ x1 + x2 = 2 ..... (1)
⇒ y1 + y2 = 4 ..... (2)
We know that lengths of the sides of the square are equal.
AB = BC = CD = DA.
We know that distance between two points (x1, y1) and (x2, y2) is
$\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}$
⇒ AB = BC
⇒ AB2 = BC2
⇒ (x1 - (- 1))2 + (y1 - 2)2 = (x1 - 3)2 + (y1 - 2)2
⇒ x12 + 1 + 2x1 + y12 + 4 - 4y1 = x12 - 6x1 + 9 + y12 + 4 - 4y1
⇒ 8x1 = 8
$\Rightarrow \mathrm{x}_{1}=\frac{8}{8}$
⇒ x1 = 1 ..... (3)
From (1)
⇒ x2 = 2 - 1 = 1 ..... (4)
We know that points ABC form right angled isosceles triangle.
We have AB2 + BC2 = AC2
⇒ 2AB2 = (- 1 - 3)2 + (2 - 2)2
⇒ 2((1 - (- 1))2 + (y1 - 2)2) = (- 4)2 + (0)2
⇒ 2(22 + (y1 - 2)2) = 8
⇒ 4 + (y1 - 2)2 = 8
⇒ (y1 - 2)2 = 4
⇒ y1 - 2 = ±2
⇒ y1 = 2 - 2 (or) y1 = 2 + 2
⇒ y1 = 0 (or) y1 = 4
From (2)
⇒ y2 = 4 - 0
⇒ y2 = 4
⇒ y2 = 4 - 4
⇒ y2 = 0
It is clear that the other two points are (1, 0) and (1, 4).
∴ The other two points are (1, 0) and (1, 4).
[Hint: Take A as the origin and AB and AD as x and y - axis respectively. Let AB = a, AD = b]
Let us assume A as the origin (0, 0) and AB and AD as x and y axis with length a and b units.
Then we get points B to be (a, 0), D to be (0, b) and C to be (a, b).
Let us assume P(x, y) be any point in a plane of the rectangle.
We need to prove PA2 + PC2 = PB2 + PD2.
We know that distance between two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Let us assume L.H.S,
⇒ PA2 + PC2 = ((x - 0)2 + (y - 0)2) + ((x - a)2 + (y - b)2)
⇒ PA2 + PC2 = x2 + y2 + x2 - 2ax + a2 + y2 - 2by + b2
⇒ PA2 + PC2 = (x2 - 2ax + a2 + y2) + (x2 + y2 - 2by + b2)
⇒ PA2 + PC2 = ((x - a)2 + (y - 0)2) + ((x - 0)2 + (y - b)2)
⇒ PA2 + PC2 = PB2 + PD2
⇒ L.H.S = R.H.S
∴ Thus proved.
Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.
We get the vertices of the rectangle as follows.
⇒ A = (0, 0)
⇒ B = (a, 0)
⇒ C = (a, b)
⇒ D = (0, b)
We need to prove the lengths of the diagonals are equal.
i.e., AC = BD
We know that distance between two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Let us find the individual lengths of diagonals,
⇒ $A C=\sqrt{(0-a)^{2}+(0-b)^{2}}$
⇒ $A C=\sqrt{a^{2}+b^{2}}$ ..... (1)
⇒ $\mathrm{BD}=\sqrt{(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}}$
⇒ $\mathrm{BD}=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$ ..... (2)
From (1) and (2), we can clearly say that AC = BD.
∴ The diagonals of a rectangle are equal.
Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.
We get the vertices of the rectangle as follows.
⇒ A = (0, 0)
⇒ B = (a, 0)
⇒ C = (a, b)
⇒ D = (0, b)
We need to prove that the sum of squares of the diagonals of a rectangle is equal to the sum of squares of its sides.
i.e., AC2 + BD2 = AB2 + BC2 + CD2 + DA2
We know that distance between two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Assume L.H.S
⇒ AC2 + BD2 = ((0 - a)2 + (0 - b)2) + ((a - 0)2 + (0 - b)2)
⇒ AC2 + BD2 = a2 + b2 + a2 + b2
⇒ AC2 + BD2 = 2(a2 + b2) ..... - (1)
Assume R.H.S
⇒ AB2 + BC2 + CD2 + DA2 = ((0 - a)2 + (0 - 0)2) + ((a - a)2 + (0 - b)2) + ((a - 0)2 + (b - b)2) + ((0 - 0)2 + (b - 0)2)
⇒ AB2 + BC2 + CD2 + DA2 = a2 + 0 + 0 + b2 + a2 + 0 + 0 + b2
⇒ AB2 + BC2 + CD2 + DA2 = 2(a2 + b2) .... (2)
From (1) and (2), we can clearly say that,
⇒ AC2 + BD2 = AB2 + BC2 + CD2 + DA2
∴ Thus proved.
⇒ AB = BC
⇒ AB2 = BC2
⇒ (x1 - (- 1))2 + (y1 - 2)2 = (x1 - 3)2 + (y1 - 2)2
⇒ x12 + 1 + 2x1 + y12 + 4 - 4y1 = x12 - 6x1 + 9 + y12 + 4 - 4y1
⇒ 8x1 = 8
$\Rightarrow \mathrm{x}_{1}=\frac{8}{8}$
⇒ x1 = 1 ..... (3)
From (1)
⇒ x2 = 2 - 1 = 1 ..... (4)
We know that points ABC form right angled isosceles triangle.
We have AB2 + BC2 = AC2
⇒ 2AB2 = (- 1 - 3)2 + (2 - 2)2
⇒ 2((1 - (- 1))2 + (y1 - 2)2) = (- 4)2 + (0)2
⇒ 2(22 + (y1 - 2)2) = 8
⇒ 4 + (y1 - 2)2 = 8
⇒ (y1 - 2)2 = 4
⇒ y1 - 2 = ±2
⇒ y1 = 2 - 2 (or) y1 = 2 + 2
⇒ y1 = 0 (or) y1 = 4
From (2)
⇒ y2 = 4 - 0
⇒ y2 = 4
⇒ y2 = 4 - 4
⇒ y2 = 0
It is clear that the other two points are (1, 0) and (1, 4).
∴ The other two points are (1, 0) and (1, 4).
Question 27
If ABCD be a rectangle and P be any point in a plane of the rectangle, then prove that
PA2 + PC2 = PB2 + PD2.
Sol :[Hint: Take A as the origin and AB and AD as x and y - axis respectively. Let AB = a, AD = b]
Let us assume A as the origin (0, 0) and AB and AD as x and y axis with length a and b units.
Then we get points B to be (a, 0), D to be (0, b) and C to be (a, b).
Let us assume P(x, y) be any point in a plane of the rectangle.
We need to prove PA2 + PC2 = PB2 + PD2.
We know that distance between two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Let us assume L.H.S,
⇒ PA2 + PC2 = ((x - 0)2 + (y - 0)2) + ((x - a)2 + (y - b)2)
⇒ PA2 + PC2 = x2 + y2 + x2 - 2ax + a2 + y2 - 2by + b2
⇒ PA2 + PC2 = (x2 - 2ax + a2 + y2) + (x2 + y2 - 2by + b2)
⇒ PA2 + PC2 = ((x - a)2 + (y - 0)2) + ((x - 0)2 + (y - b)2)
⇒ PA2 + PC2 = PB2 + PD2
⇒ L.H.S = R.H.S
∴ Thus proved.
Question 28
Prove, using co - ordinates that diagonals of a rectangle are equal.
Sol :Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.
We get the vertices of the rectangle as follows.
⇒ A = (0, 0)
⇒ B = (a, 0)
⇒ C = (a, b)
⇒ D = (0, b)
We need to prove the lengths of the diagonals are equal.
i.e., AC = BD
We know that distance between two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Let us find the individual lengths of diagonals,
⇒ $A C=\sqrt{(0-a)^{2}+(0-b)^{2}}$
⇒ $A C=\sqrt{a^{2}+b^{2}}$ ..... (1)
⇒ $\mathrm{BD}=\sqrt{(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}}$
⇒ $\mathrm{BD}=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$ ..... (2)
From (1) and (2), we can clearly say that AC = BD.
∴ The diagonals of a rectangle are equal.
Question 29
Prove, using coordinates that the sum of squares of the diagonals of a rectangle is equal to the sum of
squares of its sides.
Sol :Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.
We get the vertices of the rectangle as follows.
⇒ A = (0, 0)
⇒ B = (a, 0)
⇒ C = (a, b)
⇒ D = (0, b)
We need to prove that the sum of squares of the diagonals of a rectangle is equal to the sum of squares of its sides.
i.e., AC2 + BD2 = AB2 + BC2 + CD2 + DA2
We know that distance between two points (x1, y1) and (x2, y2) is $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$.
Assume L.H.S
⇒ AC2 + BD2 = ((0 - a)2 + (0 - b)2) + ((a - 0)2 + (0 - b)2)
⇒ AC2 + BD2 = a2 + b2 + a2 + b2
⇒ AC2 + BD2 = 2(a2 + b2) ..... - (1)
Assume R.H.S
⇒ AB2 + BC2 + CD2 + DA2 = ((0 - a)2 + (0 - 0)2) + ((a - a)2 + (0 - b)2) + ((a - 0)2 + (b - b)2) + ((0 - 0)2 + (b - 0)2)
⇒ AB2 + BC2 + CD2 + DA2 = a2 + 0 + 0 + b2 + a2 + 0 + 0 + b2
⇒ AB2 + BC2 + CD2 + DA2 = 2(a2 + b2) .... (2)
From (1) and (2), we can clearly say that,
⇒ AC2 + BD2 = AB2 + BC2 + CD2 + DA2
∴ Thus proved.
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