The distance between the points (7, -5) and (3, -1) is :
4 units
3 units
4 2 4{\sqrt2} 4 2
5 units
Answer
Let (7, -5) = (x1 , y1 ) and (3, -1) = (x2 , y2 )
Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 3 − 7 ) 2 + ( ( − 1 ) − ( − 5 ) ) 2 = ( − 4 ) 2 + ( 4 ) 2 = 16 + 16 = 32 = 4 2 units \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(3 - 7)^2 + ((-1) - (-5))^2}\\[1em] = \sqrt{(-4)^2 + (4)^2}\\[1em] = \sqrt{16 + 16}\\[1em] = \sqrt{32}\\[1em] = 4{\sqrt2} \text{units} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 3 − 7 ) 2 + (( − 1 ) − ( − 5 ) ) 2 = ( − 4 ) 2 + ( 4 ) 2 = 16 + 16 = 32 = 4 2 units
Hence, option 3 is the correct option.
The distance of point (-4, 3) from the origin is :
5 units
-5 units
4 units
3 units
Answer
Since, distance between origin and (x, y) = x 2 + y 2 \sqrt{x^2 + y^2} x 2 + y 2
∴ Distance between origin and the point (-4, 3)
= ( − 4 ) 2 + 3 2 = 16 + 9 = 25 = 5 units = \sqrt{(-4)^2 + 3^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = \text{5 units} = ( − 4 ) 2 + 3 2 = 16 + 9 = 25 = 5 units
Hence, option 1 is the correct option.
The distance between the points (-3, 2) and (x, 10) is 10 units. The value of x is :
3
-9
3 or -9
3 and -9
Answer
Let (-3, 2) = (x1 , y1 ) and (x, 10) = (x2 , y2 )
Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 10 = ( x − ( − 3 ) ) 2 + ( 10 − 2 ) 2 ⇒ 10 = ( x + 3 ) 2 + 8 2 ⇒ 10 2 = ( x + 3 ) 2 + ( 8 ) 2 ⇒ 100 = ( x + 3 ) 2 + 64 ⇒ 100 − 64 = ( x + 3 ) 2 ⇒ 36 = ( 3 + x ) 2 ⇒ 36 = 3 + x ± 6 = 3 + x 6 = x + 3 or − 6 = 3 + x x = 6 − 3 or − 6 − 3 = x x = 3 or − 9 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] 10 = \sqrt{(x - (-3))^2 + (10 - 2)^2}\\[1em] ⇒ 10 = \sqrt{(x + 3)^2 + 8^2}\\[1em] ⇒ 10^2 = (x + 3)^2 + (8)^2\\[1em] ⇒ 100 = (x + 3)^2 + 64\\[1em] ⇒ 100 - 64 = (x + 3)^2\\[1em] ⇒ 36 = (3 + x)^2\\[1em] ⇒ \sqrt{36} = 3 + x\\[1em] ± 6 = 3 + x\\[1em] 6 = x + 3 \text{ or} -6 = 3 + x \\[1em] x = 6 - 3 \text{ or} -6 -3 = x \\[1em] x = 3 \text{ or} -9\\[1em] ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 10 = ( x − ( − 3 ) ) 2 + ( 10 − 2 ) 2 ⇒ 10 = ( x + 3 ) 2 + 8 2 ⇒ 1 0 2 = ( x + 3 ) 2 + ( 8 ) 2 ⇒ 100 = ( x + 3 ) 2 + 64 ⇒ 100 − 64 = ( x + 3 ) 2 ⇒ 36 = ( 3 + x ) 2 ⇒ 36 = 3 + x ± 6 = 3 + x 6 = x + 3 or − 6 = 3 + x x = 6 − 3 or − 6 − 3 = x x = 3 or − 9
Hence, option 3 is the correct option.
The point (x, y) is equidistant from the points (3, 6) and (-3, 4); the relation between x and y is :
3x - y = 5
3x - y - 5 = 0
y - 3x = 0
3x + y = 5
Answer
Given (x, y) is equidistant from (3, 6) and (-3, 4)
i.e. distance between (x, y) and (3, 6) = distance between (x, y) and (-3, 4)
( x − 3 ) 2 + ( y − 6 ) 2 = ( x − ( − 3 ) ) 2 + ( y − 4 ) 2 ⇒ ( x − 3 ) 2 + ( y − 6 ) 2 = ( x + 3 ) 2 + ( y − 4 ) 2 ⇒ ( x − 3 ) 2 + ( y − 6 ) 2 = ( x + 3 ) 2 + ( y − 4 ) 2 ⇒ x 2 + 9 − 6 x + y 2 + 36 − 12 y = x 2 + 9 + 6 x + y 2 + 16 − 8 y ⇒ 36 − 16 = 6 x + 6 x + 12 y − 8 y ⇒ 20 = 12 x + 4 y ⇒ 3 x + y = 5 \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x - (-3))^2 + (y - 4)^2}\\[1em] ⇒ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2}\\[1em] ⇒ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2\\[1em] ⇒ x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 -8y\\[1em] ⇒ 36 - 16 = 6x + 6x + 12y -8y\\[1em] ⇒ 20 = 12x + 4y\\[1em] ⇒ 3x + y = 5 ( x − 3 ) 2 + ( y − 6 ) 2 = ( x − ( − 3 ) ) 2 + ( y − 4 ) 2 ⇒ ( x − 3 ) 2 + ( y − 6 ) 2 = ( x + 3 ) 2 + ( y − 4 ) 2 ⇒ ( x − 3 ) 2 + ( y − 6 ) 2 = ( x + 3 ) 2 + ( y − 4 ) 2 ⇒ x 2 + 9 − 6 x + y 2 + 36 − 12 y = x 2 + 9 + 6 x + y 2 + 16 − 8 y ⇒ 36 − 16 = 6 x + 6 x + 12 y − 8 y ⇒ 20 = 12 x + 4 y ⇒ 3 x + y = 5
Hence, option 4 is the correct option.
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is :
12 units
6 units
5 units
10 units
Answer
Let (0, 4) = (x1 , y1 ), (0, 0) = (x2 , y2 ) and (3, 0) = (x3 , y3 )
Distance between the given points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
The perimeter of a triangle
= ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( x 3 − x 2 ) 2 + ( y 3 − y 2 ) 2 + ( x 3 − x 1 ) 2 + ( y 3 − y 1 ) 2 = ( 0 − 0 ) 2 + ( 0 − 4 ) 2 + ( 3 − 0 ) 2 + ( 0 − 0 ) 2 + ( 3 − 0 ) 2 + ( 0 − 4 ) 2 = ( − 4 ) 2 + ( 3 ) 2 + ( 3 ) 2 + ( − 4 ) 2 = 16 + 9 + 9 + 16 = 4 + 3 + 25 = 4 + 3 + 5 = 12 units = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} + \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} + \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}\\[1em] = \sqrt{(0 - 0)^2 + (0 - 4)^2} + \sqrt{(3 - 0)^2 + (0 - 0)^2} + \sqrt{(3 - 0)^2 + (0 - 4)^2}\\[1em] = \sqrt{(- 4)^2} + \sqrt{(3)^2} + \sqrt{(3)^2 + (- 4)^2}\\[1em] = \sqrt{16} + \sqrt{9} + \sqrt{9 + 16}\\[1em] = 4 + 3 + \sqrt{25}\\[1em] = 4 + 3 + 5\\[1em] = 12 \text{units} = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( x 3 − x 2 ) 2 + ( y 3 − y 2 ) 2 + ( x 3 − x 1 ) 2 + ( y 3 − y 1 ) 2 = ( 0 − 0 ) 2 + ( 0 − 4 ) 2 + ( 3 − 0 ) 2 + ( 0 − 0 ) 2 + ( 3 − 0 ) 2 + ( 0 − 4 ) 2 = ( − 4 ) 2 + ( 3 ) 2 + ( 3 ) 2 + ( − 4 ) 2 = 16 + 9 + 9 + 16 = 4 + 3 + 25 = 4 + 3 + 5 = 12 units
Hence, option 1 is the correct option.
Find the distance between the following pairs of points :
(i) (-3, 6) and (2, -6)
(ii) (-a, -b) and (a, b)
(iii) ( 3 5 , 2 ) \Big(\dfrac{3}{5},2\Big) ( 5 3 , 2 ) ( − 1 5 , 1 2 5 ) \Big(-\dfrac{1}{5}, 1\dfrac{2}{5}\Big) ( − 5 1 , 1 5 2 )
(iv) ( 3 + 1 , 1 ) \Big({\sqrt3 +1},1\Big) ( 3 + 1 , 1 ) 3 \sqrt{3} 3
Answer
(i) Let (-3, 6) = (x1 , y1 ) and (2, -6) = (x2 , y2 )
⇒ Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 2 − ( − 3 ) ) 2 + ( − 6 − 6 ) 2 = ( 5 ) 2 + ( − 12 ) 2 = 25 + 144 = 169 = 13 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(2 - (-3))^2 + (-6 - 6)^2}\\[1em] = \sqrt{(5)^2 + (-12)^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = 13 ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 2 − ( − 3 ) ) 2 + ( − 6 − 6 ) 2 = ( 5 ) 2 + ( − 12 ) 2 = 25 + 144 = 169 = 13
Hence, distance between the given points is 13.
(ii) Let (-a, -b) = (x1 , y1 ) and (a, b) = (x2 , y2 )
⇒ Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( a − ( − a ) ) 2 + ( b − ( − b ) ) 2 = ( 2 a ) 2 + ( 2 b ) 2 = 4 a 2 + 4 b 2 = 2 a 2 + b 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(a - (-a))^2 + (b - (-b))^2}\\[1em] = \sqrt{(2a)^2 + (2b)^2}\\[1em] = \sqrt{4a^2 + 4b^2}\\[1em] = 2\sqrt{a^2 + b^2}\\[1em] ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( a − ( − a ) ) 2 + ( b − ( − b ) ) 2 = ( 2 a ) 2 + ( 2 b ) 2 = 4 a 2 + 4 b 2 = 2 a 2 + b 2
Hence, distance between the given points is 2 a 2 + b 2 2\sqrt{a^2 + b^2} 2 a 2 + b 2
(iii) Let ( 3 5 , 2 ) \Big(\dfrac{3}{5},2\Big) ( 5 3 , 2 ) 1 , y1 ) and ( − 1 5 , 1 2 5 ) \Big(-\dfrac{1}{5}, 1\dfrac{2}{5}\Big) ( − 5 1 , 1 5 2 ) 2 , y2 )
⇒ Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( − 1 5 − 3 5 ) 2 + ( ( 1 2 5 − 2 ) ) 2 = ( − 4 5 ) 2 + ( 7 − 10 5 ) 2 = 16 25 + ( − 3 5 ) 2 = 16 25 + 9 25 = 25 25 = 1 = 1 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{\Big(\dfrac{-1}{5} - \dfrac{3}{5} \Big)^2 + \Big(\Big(1\dfrac{2}{5} - 2 \Big)\Big)^2}\\[1em] = \sqrt{\Big(\dfrac{-4}{5}\Big)^2 + \Big(\dfrac{7-10}{5}\Big)^2}\\[1em] = \sqrt{\dfrac{16}{25} + \Big(\dfrac{-3}{5}\Big)^2}\\[1em] = \sqrt{\dfrac{16}{25} + \dfrac{9}{25}}\\[1em] = \sqrt{\dfrac{25}{25}}\\[1em] = \sqrt{1}\\[1em] = 1 ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 5 − 1 − 5 3 ) 2 + ( ( 1 5 2 − 2 ) ) 2 = ( 5 − 4 ) 2 + ( 5 7 − 10 ) 2 = 25 16 + ( 5 − 3 ) 2 = 25 16 + 25 9 = 25 25 = 1 = 1
Hence, distance between the given points is 1.
(iv) Let ( 3 + 1 , 1 ) \Big({\sqrt3 +1},1\Big) ( 3 + 1 , 1 ) 1 , y1 ) and (0, 3 \sqrt{3} 3 2 , y2 )
⇒ Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 0 − ( 3 + 1 ) ) 2 + ( 3 − 1 ) 2 = ( − ( 3 + 1 ) ) 2 + ( 3 − 1 ) 2 = 3 + 1 + 2 3 + 1 + 3 − 2 3 = 8 = 2 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(0 - ({\sqrt3 +1}))^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{(-{(\sqrt3 +1)})^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{3 + 1 + 2\sqrt3 + 1 + 3 - 2\sqrt3}\\[1em] = \sqrt{8}\\[1em] = 2\sqrt2 ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 0 − ( 3 + 1 ) ) 2 + ( 3 − 1 ) 2 = ( − ( 3 + 1 ) ) 2 + ( 3 − 1 ) 2 = 3 + 1 + 2 3 + 1 + 3 − 2 3 = 8 = 2 2
Hence, distance between the given points is 2 2 2\sqrt2 2 2
Find the distance between the origin and the point :
(i) (-8, 6)
(ii) (-5, -12)
(iii) (8, -15)
Answer
(i) Since, distance between origin and (x, y) = x 2 + y 2 \sqrt{x^2 + y^2} x 2 + y 2
∴ Distance between origin and the point (-8, 6)
= ( − 8 ) 2 + 6 2 = 64 + 36 = 100 = 10 = \sqrt{(-8)^2 + 6^2}\\[1em] = \sqrt{64 + 36}\\[1em] = \sqrt{100}\\[1em] = \text{10} = ( − 8 ) 2 + 6 2 = 64 + 36 = 100 = 10
Hence, the distance between the origin and the point (-8, 6) is 10.
(ii) Since, distance between origin and (x, y) = x 2 + y 2 \sqrt{x^2 + y^2} x 2 + y 2
∴ Distance between origin and the point (-5, -12)
= ( − 5 ) 2 + ( − 12 ) 2 = 25 + 144 = 169 = 13 = \sqrt{(-5)^2 + (-12)^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = \text{13} = ( − 5 ) 2 + ( − 12 ) 2 = 25 + 144 = 169 = 13
Hence, the distance between the origin and the point (-5, -12) is 13.
(iii) Since, distance between origin and (x, y) = x 2 + y 2 \sqrt{x^2 + y^2} x 2 + y 2
∴ Distance between origin and the point (8, -15)
= 8 2 + ( − 15 ) 2 = 64 + 225 = 289 = 17 = \sqrt{8^2 + (-15)^2}\\[1em] = \sqrt{64 + 225}\\[1em] = \sqrt{289}\\[1em] = \text{17} = 8 2 + ( − 15 ) 2 = 64 + 225 = 289 = 17
Hence, the distance between the origin and the point is 17.
The distance between the points (3, 1) and (0, x) is 5. Find x.
Answer
Let (3, 1) = (x1 , y1 ) and (0, x) = (x2 , y2 )
⇒ Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ⇒ 5 = ( 0 − 3 ) 2 + ( x − 1 ) 2 ⇒ 5 = ( − 3 ) 2 + ( x − 1 ) 2 ⇒ 5 2 = ( − 3 ) 2 + ( x − 1 ) 2 ⇒ 25 = 9 + ( x − 1 ) 2 ⇒ 25 − 9 = ( x − 1 ) 2 ⇒ 16 = ( x − 1 ) 2 ⇒ 16 = x − 1 ⇒ 4 = x − 1 and − 4 = x − 1 ⇒ x = 4 + 1 and x = − 4 + 1 ⇒ x = 5 and − 3 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 5 = \sqrt{(0 - 3)^2 + (x - 1)^2}\\[1em] ⇒ 5 = \sqrt{(-3)^2 + (x - 1)^2}\\[1em] ⇒ 5^2 = (-3)^2 + (x - 1)^2\\[1em] ⇒ 25 = 9 + (x - 1)^2\\[1em] ⇒ 25 - 9 = (x - 1)^2\\[1em] ⇒ 16 = (x - 1)^2\\[1em] ⇒ \sqrt{16} = x - 1\\[1em] ⇒ 4 = x - 1 \text{ and} -4 = x -1\\[1em] ⇒ x = 4 + 1 \text{ and } x = -4 + 1\\[1em] ⇒ x = 5 \text{ and} -3\\[1em] ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ⇒ 5 = ( 0 − 3 ) 2 + ( x − 1 ) 2 ⇒ 5 = ( − 3 ) 2 + ( x − 1 ) 2 ⇒ 5 2 = ( − 3 ) 2 + ( x − 1 ) 2 ⇒ 25 = 9 + ( x − 1 ) 2 ⇒ 25 − 9 = ( x − 1 ) 2 ⇒ 16 = ( x − 1 ) 2 ⇒ 16 = x − 1 ⇒ 4 = x − 1 and − 4 = x − 1 ⇒ x = 4 + 1 and x = − 4 + 1 ⇒ x = 5 and − 3
Hence, the value of x = 5 and -3.
Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).
Answer
Let the co-ordinates of the point on the x-axis be (x, 0).
Since, distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Let (x, 0) = (x1 , y1 ) and (11, -8) = (x2 , y2 )
⇒ Distance between the given points =
⇒ 17 = ( 11 − x ) 2 + ( − 8 − 0 ) 2 ⇒ 17 = ( 11 − x ) 2 + ( − 8 ) 2 ⇒ 17 2 = ( 11 − x ) 2 + ( − 8 ) 2 ⇒ 289 = 121 + x 2 − 22 x + 64 ⇒ 289 = 185 + x 2 − 22 x ⇒ 185 + x 2 − 22 x − 289 = 0 ⇒ x 2 − ( 26 x − 4 x ) − 104 = 0 ⇒ x 2 − 26 x + 4 x − 104 = 0 ⇒ ( x 2 − 26 x ) + ( 4 x − 104 ) = 0 ⇒ x ( x − 26 ) + 4 ( x − 26 ) = 0 ⇒ ( x − 26 ) ( x + 4 ) = 0 ⇒ x = 26 and − 4 \\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8 - 0)^2}\\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8)^2}\\[1em] ⇒ 17^2 = (11 - x)^2 + (-8)^2\\[1em] ⇒ 289 = 121 + x^2 - 22x + 64\\[1em] ⇒ 289 = 185 + x^2 - 22x\\[1em] ⇒ 185 + x^2 - 22x - 289 = 0\\[1em] ⇒ x^2 - (26x - 4x) - 104 = 0\\[1em] ⇒ x^2 - 26x + 4x - 104 = 0\\[1em] ⇒ (x^2 - 26x) + (4x - 104) = 0\\[1em] ⇒ x(x - 26) + 4(x - 26) = 0\\[1em] ⇒ (x - 26)(x + 4) = 0\\[1em] ⇒ x = 26 \text{ and} -4 ⇒ 17 = ( 11 − x ) 2 + ( − 8 − 0 ) 2 ⇒ 17 = ( 11 − x ) 2 + ( − 8 ) 2 ⇒ 1 7 2 = ( 11 − x ) 2 + ( − 8 ) 2 ⇒ 289 = 121 + x 2 − 22 x + 64 ⇒ 289 = 185 + x 2 − 22 x ⇒ 185 + x 2 − 22 x − 289 = 0 ⇒ x 2 − ( 26 x − 4 x ) − 104 = 0 ⇒ x 2 − 26 x + 4 x − 104 = 0 ⇒ ( x 2 − 26 x ) + ( 4 x − 104 ) = 0 ⇒ x ( x − 26 ) + 4 ( x − 26 ) = 0 ⇒ ( x − 26 ) ( x + 4 ) = 0 ⇒ x = 26 and − 4
Hence, the co-ordinates of points are (26, 0) and (-4, 0).
Find the co-ordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).
Answer
Let (0, y) = (x1 , y1 ) and (-8, 4) = (x2 , y2 )
⇒ Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ⇒ 10 = ( − 8 − 0 ) 2 + ( 4 − y ) 2 ⇒ 10 = ( − 8 ) 2 + ( 4 − y ) 2 ⇒ 10 2 = 64 + 16 + y 2 − 8 y ⇒ 100 = 80 + y 2 − 8 y ⇒ 80 + y 2 − 8 y − 100 = 0 ⇒ y 2 − 8 y − 20 = 0 ⇒ y 2 − ( 10 y − 2 y ) − 20 = 0 ⇒ y 2 − 10 y + 2 y − 20 = 0 ⇒ ( y 2 − 10 y ) + ( 2 y − 20 ) = 0 ⇒ y ( y − 10 ) + 2 ( y − 10 ) = 0 ⇒ ( y − 10 ) ( y + 2 ) = 0 ⇒ y = 10 and − 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 10 = \sqrt{(-8 - 0)^2 + (4 - y)^2}\\[1em] ⇒ 10 = \sqrt{(-8)^2 + (4 - y)^2}\\[1em] ⇒ 10^2 = 64 + 16 + y^2 - 8y\\[1em] ⇒ 100 = 80 + y^2 - 8y\\[1em] ⇒ 80 + y^2 - 8y - 100 = 0 \\[1em] ⇒ y^2 - 8y - 20 = 0 \\[1em] ⇒ y^2 - (10y - 2y) - 20 = 0 \\[1em] ⇒ y^2 - 10y + 2y - 20 = 0 \\[1em] ⇒ (y^2 - 10y) + (2y - 20) = 0 \\[1em] ⇒ y(y - 10) + 2(y - 10) = 0 \\[1em] ⇒ (y - 10)(y + 2) = 0 \\[1em] ⇒ y = 10 \text{ and } -2 ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ⇒ 10 = ( − 8 − 0 ) 2 + ( 4 − y ) 2 ⇒ 10 = ( − 8 ) 2 + ( 4 − y ) 2 ⇒ 1 0 2 = 64 + 16 + y 2 − 8 y ⇒ 100 = 80 + y 2 − 8 y ⇒ 80 + y 2 − 8 y − 100 = 0 ⇒ y 2 − 8 y − 20 = 0 ⇒ y 2 − ( 10 y − 2 y ) − 20 = 0 ⇒ y 2 − 10 y + 2 y − 20 = 0 ⇒ ( y 2 − 10 y ) + ( 2 y − 20 ) = 0 ⇒ y ( y − 10 ) + 2 ( y − 10 ) = 0 ⇒ ( y − 10 ) ( y + 2 ) = 0 ⇒ y = 10 and − 2
Hence, the co-ordinates of points are (0, 10) and (0, -2).
A point A is at a distance of 10 \sqrt{10} 10
Answer
Let (a, 2a) = (x1 , y1 ) and (4, 3) = (x2 , y2 )
⇒ Distance between the given points =
( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ⇒ 10 = ( 4 − a ) 2 + ( 3 − 2 a ) 2 ⇒ 10 = ( 4 − a ) 2 + ( 3 − 2 a ) 2 ⇒ 10 = 16 + a 2 − 8 a + 9 + 4 a 2 − 12 a ⇒ 10 = 25 + 5 a 2 − 20 a ⇒ 25 + 5 a 2 − 20 a − 10 = 0 ⇒ 5 a 2 − 20 a + 15 = 0 ⇒ a 2 − 4 a + 3 = 0 ⇒ a 2 − ( 3 a + 1 a ) + 3 = 0 ⇒ a 2 − 3 a − 1 a + 3 = 0 ⇒ a ( a − 3 ) − 1 ( a − 3 ) = 0 ⇒ ( a − 3 ) ( a − 1 ) = 0 ⇒ a = 3 and 1 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ \sqrt{10} = \sqrt{(4 - a)^2 + (3 - 2a)^2}\\[1em] ⇒ 10 = (4 - a)^2 + (3 - 2a)^2\\[1em] ⇒ 10 = 16 + a^2 - 8a + 9 + 4a^2 - 12a\\[1em] ⇒ 10 = 25 + 5a^2 - 20a\\[1em] ⇒ 25 + 5a^2 - 20a - 10 = 0\\[1em] ⇒ 5a^2 - 20a + 15 = 0\\[1em] ⇒ a^2 - 4a + 3 = 0\\[1em] ⇒ a^2 - (3a + 1a) + 3 = 0\\[1em] ⇒ a^2 - 3a - 1a + 3 = 0\\[1em] ⇒ a(a - 3) - 1(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 1) = 0\\[1em] ⇒ a = 3 \text{ and } 1 ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ⇒ 10 = ( 4 − a ) 2 + ( 3 − 2 a ) 2 ⇒ 10 = ( 4 − a ) 2 + ( 3 − 2 a ) 2 ⇒ 10 = 16 + a 2 − 8 a + 9 + 4 a 2 − 12 a ⇒ 10 = 25 + 5 a 2 − 20 a ⇒ 25 + 5 a 2 − 20 a − 10 = 0 ⇒ 5 a 2 − 20 a + 15 = 0 ⇒ a 2 − 4 a + 3 = 0 ⇒ a 2 − ( 3 a + 1 a ) + 3 = 0 ⇒ a 2 − 3 a − 1 a + 3 = 0 ⇒ a ( a − 3 ) − 1 ( a − 3 ) = 0 ⇒ ( a − 3 ) ( a − 1 ) = 0 ⇒ a = 3 and 1
For each value of a = x, we can find the corresponding value of y:
If a = 3, then y = 2a = 6 If a = 1, then y = 2a = 2 Hence, the co-ordinates of point A are (3, 6) and (1, 2).
A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
Answer
Given point (2, -1) is equidistant from (a, 7) and (-3, a).
i.e. distance between (2, -1) and (a, 7) = distance between (2, -1) and (-3, a)
( a − 2 ) 2 + ( 7 − ( − 1 ) ) 2 = ( ( − 3 ) − 2 ) 2 + ( a − ( − 1 ) ) 2 ⇒ ( a − 2 ) 2 + ( 7 + 1 ) 2 = ( − 3 − 2 ) 2 + ( a + 1 ) 2 ⇒ ( a − 2 ) 2 + ( 8 ) 2 = ( − 5 ) 2 + ( a + 1 ) 2 ⇒ ( a − 2 ) 2 + ( 8 ) 2 = ( − 5 ) 2 + ( a + 1 ) 2 ⇒ a 2 + 4 − 4 a + 64 = 25 + a 2 + 1 + 2 a ⇒ a 2 − 4 a + 68 = a 2 + 2 a + 26 ⇒ − 4 a + 68 = 2 a + 26 ⇒ 68 − 26 = 2 a + 4 a ⇒ 6 a = 42 ⇒ a = 42 6 ⇒ a = 7 \sqrt{(a - 2)^2 + (7 - (-1))^2} = \sqrt{((-3) - 2)^2 + (a - (-1))^2}\\[1em] ⇒ \sqrt{(a - 2)^2 + (7 + 1)^2} = \sqrt{(-3 - 2)^2 + (a + 1)^2}\\[1em] ⇒ \sqrt{(a - 2)^2 + (8)^2} = \sqrt{(-5)^2 + (a + 1)^2}\\[1em] ⇒ (a - 2)^2 + (8)^2 = (-5)^2 + (a + 1)^2\\[1em] ⇒ a^2 + 4 - 4a + 64 = 25 + a^2 + 1 + 2a\\[1em] ⇒ a^2 - 4a + 68 = a^2 + 2a + 26\\[1em] ⇒ - 4a + 68 = 2a + 26\\[1em] ⇒ 68 - 26 = 2a + 4a\\[1em] ⇒ 6a = 42\\[1em] ⇒ a = \dfrac{42}{6}\\[1em] ⇒ a = 7 ( a − 2 ) 2 + ( 7 − ( − 1 ) ) 2 = (( − 3 ) − 2 ) 2 + ( a − ( − 1 ) ) 2 ⇒ ( a − 2 ) 2 + ( 7 + 1 ) 2 = ( − 3 − 2 ) 2 + ( a + 1 ) 2 ⇒ ( a − 2 ) 2 + ( 8 ) 2 = ( − 5 ) 2 + ( a + 1 ) 2 ⇒ ( a − 2 ) 2 + ( 8 ) 2 = ( − 5 ) 2 + ( a + 1 ) 2 ⇒ a 2 + 4 − 4 a + 64 = 25 + a 2 + 1 + 2 a ⇒ a 2 − 4 a + 68 = a 2 + 2 a + 26 ⇒ − 4 a + 68 = 2 a + 26 ⇒ 68 − 26 = 2 a + 4 a ⇒ 6 a = 42 ⇒ a = 6 42 ⇒ a = 7
Hence, the value of a = 7.
What point on the x-axis is equidistant from the points (7, 6) and (-3, 4) ?
Answer
Let the required point on the x-axis be (x, 0).
Given (x, 0) is equidistant from (7, 6) and (-3, 4).
i.e. distance between (x, 0) and (7, 6) = distance between (x, 0) and (-3, 4)
( 7 − x ) 2 + ( 6 − 0 ) 2 = ( ( − 3 ) − x ) 2 + ( 4 − 0 ) 2 ⇒ ( 7 − x ) 2 + ( 6 ) 2 = ( − 3 − x ) 2 + ( 4 ) 2 ⇒ ( 7 − x ) 2 + ( 6 ) 2 = ( − 3 − x ) 2 + ( 4 ) 2 ⇒ 49 + x 2 − 14 x + 36 = 9 + x 2 + 6 x + 16 ⇒ x 2 − 14 x + 85 = x 2 + 6 x + 25 ⇒ − 14 x + 85 = 6 x + 25 ⇒ − 14 x − 6 x = 25 − 85 ⇒ − 20 x = − 60 ⇒ x = 60 20 ⇒ x = 3 \sqrt{(7 - x)^2 + (6 - 0)^2} = \sqrt{((-3) - x)^2 + (4 - 0)^2}\\[1em] ⇒ \sqrt{(7 - x)^2 + (6)^2} = \sqrt{(-3 - x)^2 + (4)^2}\\[1em] ⇒ (7 - x)^2 + (6)^2 = (-3 - x)^2 + (4)^2\\[1em] ⇒ 49 + x^2 - 14x + 36 = 9 + x^2 + 6x + 16\\[1em] ⇒ x^2 - 14x + 85 = x^2 + 6x + 25\\[1em] ⇒ - 14x + 85 = 6x + 25\\[1em] ⇒ - 14x - 6x = 25 - 85\\[1em] ⇒ - 20x = -60\\[1em] ⇒ x = \dfrac{60}{20}\\[1em] ⇒ x = 3 ( 7 − x ) 2 + ( 6 − 0 ) 2 = (( − 3 ) − x ) 2 + ( 4 − 0 ) 2 ⇒ ( 7 − x ) 2 + ( 6 ) 2 = ( − 3 − x ) 2 + ( 4 ) 2 ⇒ ( 7 − x ) 2 + ( 6 ) 2 = ( − 3 − x ) 2 + ( 4 ) 2 ⇒ 49 + x 2 − 14 x + 36 = 9 + x 2 + 6 x + 16 ⇒ x 2 − 14 x + 85 = x 2 + 6 x + 25 ⇒ − 14 x + 85 = 6 x + 25 ⇒ − 14 x − 6 x = 25 − 85 ⇒ − 20 x = − 60 ⇒ x = 20 60 ⇒ x = 3
Hence, the point on the x-axis which is equidistant from the points (7, 6) and (-3, 4) is (3, 0).
Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).
Answer
Let the required point on the y-axis be (0, y).
Given (0, y) is equidistant from (5, 2) and (-4, 3).
i.e. distance between (0, y) and (5, 2) = distance between (0, y) and (-4, 3)
( 5 − 0 ) 2 + ( 2 − y ) 2 = ( ( − 4 ) − 0 ) 2 + ( 3 − y ) 2 ⇒ ( 5 ) 2 + ( 2 − y ) 2 = ( − 4 ) 2 + ( 3 − y ) 2 ⇒ ( 5 ) 2 + ( 2 − y ) 2 = ( − 4 ) 2 + ( 3 − y ) 2 ⇒ 25 + 4 + y 2 − 4 y = 16 + 9 + y 2 − 6 y ⇒ 29 + y 2 − 4 y = 25 + y 2 − 6 y ⇒ 29 − 4 y = 25 − 6 y ⇒ − 4 y + 6 y = 25 − 29 ⇒ 2 y = − 4 ⇒ y = − 4 2 ⇒ y = − 2 \sqrt{(5 - 0)^2 + (2 - y)^2} = \sqrt{((-4) - 0)^2 + (3 - y)^2}\\[1em] ⇒ \sqrt{(5)^2 + (2 - y)^2} = \sqrt{(-4)^2 + (3 - y)^2}\\[1em] ⇒ (5)^2 + (2 - y)^2 = (-4)^2 + (3 - y)^2\\[1em] ⇒ 25 + 4 + y^2 - 4y = 16 + 9 + y^2 - 6y\\[1em] ⇒ 29 + y^2 - 4y = 25 + y^2 - 6y\\[1em] ⇒ 29 - 4y = 25 - 6y\\[1em] ⇒ - 4y + 6y = 25 - 29\\[1em] ⇒ 2y = - 4\\[1em] ⇒ y = - \dfrac{4}{2}\\[1em] ⇒ y = - 2 ( 5 − 0 ) 2 + ( 2 − y ) 2 = (( − 4 ) − 0 ) 2 + ( 3 − y ) 2 ⇒ ( 5 ) 2 + ( 2 − y ) 2 = ( − 4 ) 2 + ( 3 − y ) 2 ⇒ ( 5 ) 2 + ( 2 − y ) 2 = ( − 4 ) 2 + ( 3 − y ) 2 ⇒ 25 + 4 + y 2 − 4 y = 16 + 9 + y 2 − 6 y ⇒ 29 + y 2 − 4 y = 25 + y 2 − 6 y ⇒ 29 − 4 y = 25 − 6 y ⇒ − 4 y + 6 y = 25 − 29 ⇒ 2 y = − 4 ⇒ y = − 2 4 ⇒ y = − 2
Hence, the point on the y-axis which is equidistant from the points (5, 2) and (-4, 3) is (0, -2).
A point P lies on the x-axis and another point Q lies on the y-axis.
(i) Write the ordinate of point P.
(ii) Write the abscissa of point Q.
(iii) If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.
Answer
(i) The point P lies on the x-axis, so its co-ordinates are (x, 0).
The ordinate of point P is 0.
(ii) The point Q lies on the y-axis, so its co-ordinates are (0, y).
The abscissa of point Q is 0.
(iii) If the abscissa of point P is -12, then P = (-12, 0).
And if the ordinate of point Q is -16. So, then Q = (0, -16).
The length of the line segment PQ
= ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 0 − ( − 12 ) ) 2 + ( ( − 16 ) − 0 ) 2 = ( − 12 ) 2 + ( − 16 ) 2 = 144 + 256 = 400 = 20 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(0 - (-12))^2 + ((-16) - 0)^2}\\[1em] = \sqrt{(-12)^2 + (-16)^2}\\[1em] = \sqrt{144 + 256}\\[1em] = \sqrt{400}\\[1em] = 20 = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 0 − ( − 12 ) ) 2 + (( − 16 ) − 0 ) 2 = ( − 12 ) 2 + ( − 16 ) 2 = 144 + 256 = 400 = 20
Hence, the length of PQ = 20.
Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.
Answer
Distance between the points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
The length of PQ
= ( 5 − 0 ) 2 + ( 10 − 5 ) 2 = 5 2 + 5 2 = 25 + 25 = 50 = 5 2 = \sqrt{(5 - 0)^2 + (10 - 5)^2}\\[1em] = \sqrt{5^2 + 5^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em] = ( 5 − 0 ) 2 + ( 10 − 5 ) 2 = 5 2 + 5 2 = 25 + 25 = 50 = 5 2
The length of QR
= ( 6 − 5 ) 2 + ( 3 − 10 ) 2 = 1 2 + ( − 7 ) 2 = 1 + 49 = 50 = 5 2 = \sqrt{(6 - 5)^2 + (3 - 10)^2}\\[1em] = \sqrt{1^2 + (-7)^2}\\[1em] = \sqrt{1 + 49}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em] = ( 6 − 5 ) 2 + ( 3 − 10 ) 2 = 1 2 + ( − 7 ) 2 = 1 + 49 = 50 = 5 2
The length of RP
= ( 6 − 0 ) 2 + ( 3 − 5 ) 2 = 6 2 + ( − 2 ) 2 = 36 + 4 = 40 = 2 10 = \sqrt{(6 - 0)^2 + (3 - 5)^2}\\[1em] = \sqrt{6^2 + (-2)^2}\\[1em] = \sqrt{36 + 4}\\[1em] = \sqrt{40}\\[1em] = 2\sqrt{10}\\[1em] = ( 6 − 0 ) 2 + ( 3 − 5 ) 2 = 6 2 + ( − 2 ) 2 = 36 + 4 = 40 = 2 10
PQ = QR ⇒ the triangle is isosceles triangle
Hence, the triangle PQR is an isosceles triangle.
Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
Answer
Distance between the points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
The length of PQ
= ( 6 − 0 ) 2 + ( 2 − ( − 4 ) ) 2 = 6 2 + 6 2 = 36 + 36 = 72 = 6 2 = \sqrt{(6 - 0)^2 + (2 - (-4))^2}\\[1em] = \sqrt{6^2 + 6^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em] = ( 6 − 0 ) 2 + ( 2 − ( − 4 ) ) 2 = 6 2 + 6 2 = 36 + 36 = 72 = 6 2
The length of RS
= ( − 3 − 3 ) 2 + ( − 1 − 5 ) 2 = − 6 2 + ( − 6 ) 2 = 36 + 36 = 72 = 6 2 = \sqrt{(-3 - 3)^2 + (-1 - 5)^2}\\[1em] = \sqrt{-6^2 + (-6)^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em] = ( − 3 − 3 ) 2 + ( − 1 − 5 ) 2 = − 6 2 + ( − 6 ) 2 = 36 + 36 = 72 = 6 2
The length of QR
= ( 3 − 6 ) 2 + ( 5 − 2 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 = \sqrt{(3 - 6)^2 + (5 - 2)^2}\\[1em] = \sqrt{(-3)^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em] = ( 3 − 6 ) 2 + ( 5 − 2 ) 2 = ( − 3 ) 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2
The length of SP
= ( − 3 − 0 ) 2 + ( − 1 − ( − 4 ) ) 2 = − 3 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2 = \sqrt{(-3 - 0)^2 + (-1 - (-4))^2}\\[1em] = \sqrt{-3^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em] = ( − 3 − 0 ) 2 + ( − 1 − ( − 4 ) ) 2 = − 3 2 + ( − 3 ) 2 = 9 + 9 = 18 = 3 2
PQ = RS
The length of diagonal QS =
= ( − 3 − 6 ) 2 + ( − 1 − 2 ) 2 = ( − 9 ) 2 + ( − 3 ) 2 = 81 + 9 = 90 = 3 10 = \sqrt{(-3 - 6)^2 + (-1 - 2)^2}\\[1em] = \sqrt{(-9)^2 + (-3)^2}\\[1em] = \sqrt{81 + 9}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em] = ( − 3 − 6 ) 2 + ( − 1 − 2 ) 2 = ( − 9 ) 2 + ( − 3 ) 2 = 81 + 9 = 90 = 3 10
The length of diagonal PR =
= ( 3 − 0 ) 2 + ( 5 − ( − 4 ) ) 2 = 3 2 + 9 2 = 9 + 81 = 90 = 3 10 = \sqrt{(3 - 0)^2 + (5 - (-4))^2}\\[1em] = \sqrt{3^2 + 9^2}\\[1em] = \sqrt{9 + 81}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em] = ( 3 − 0 ) 2 + ( 5 − ( − 4 ) ) 2 = 3 2 + 9 2 = 9 + 81 = 90 = 3 10
So, QS = PR
Since opposite sides are equal, and the diagonals are equal, we can conclude that the quadrilateral PQRS is a rectangle.
Hence, the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.
Answer
Distance between the given points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
The length of AB
= ( − 3 − 1 ) 2 + ( 0 − ( − 3 ) ) 2 = ( − 4 ) 2 + 3 2 = 16 + 9 = 25 = 5 = \sqrt{(-3 - 1)^2 + (0 - (-3))^2}\\[1em] = \sqrt{(-4)^2 + 3^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5 = ( − 3 − 1 ) 2 + ( 0 − ( − 3 ) ) 2 = ( − 4 ) 2 + 3 2 = 16 + 9 = 25 = 5
The length of BC
= ( 4 − ( − 3 ) ) 2 + ( 1 − 0 ) 2 = ( − 7 ) 2 + 1 2 = 49 + 1 = 50 = 5 2 = \sqrt{(4 - (-3))^2 + (1 - 0)^2}\\[1em] = \sqrt{(-7)^2 + 1^2}\\[1em] = \sqrt{49 + 1}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em] = ( 4 − ( − 3 ) ) 2 + ( 1 − 0 ) 2 = ( − 7 ) 2 + 1 2 = 49 + 1 = 50 = 5 2
The length of CA
= ( 4 − 1 ) 2 + ( 1 − ( − 3 ) ) 2 = 3 2 + 4 2 = 9 + 16 = 25 = 5 = \sqrt{(4 - 1)^2 + (1 - (-3))^2}\\[1em] = \sqrt{3^2 + 4^2}\\[1em] = \sqrt{9 + 16}\\[1em] = \sqrt{25}\\[1em] = 5 = ( 4 − 1 ) 2 + ( 1 − ( − 3 ) ) 2 = 3 2 + 4 2 = 9 + 16 = 25 = 5
If ABC is an right angled triangle,
AB2 + CA2 = ( 2 5 ) (2\sqrt5) ( 2 5 ) 2 = 52 + 52 = 25 + 25 = 50 ⇒ BC2
BC2 = AB2 + CA2 ⇒ the triangle is right angled triangle.
and,
AB = CA ⇒ the triangle is isosceles triangle.
Base of triangle = Height of the triangle = 5 units.
Area of triangle ABC = 1 2 \dfrac{1}{2} 2 1
= 1 2 × 5 × 5 = 25 2 = 12.5 sq. units = \dfrac{1}{2} \times 5 \times 5\\[1em] = \dfrac{25}{2}\\[1em] = 12.5 \text{ sq. units} = 2 1 × 5 × 5 = 2 25 = 12.5 sq. units
Hence, the triangle ABC is an isosceles right-angled triangle and area of the triangle = 12.5 sq. units.
Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.
Answer
Distance between the given points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
The length of AB
= ( 1 − 5 ) 2 + ( 5 − 6 ) 2 = ( − 4 ) 2 + ( − 1 ) 2 = 16 + 1 = 17 = \sqrt{(1 - 5)^2 + (5 - 6)^2}\\[1em] = \sqrt{(-4)^2 + (-1)^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17} = ( 1 − 5 ) 2 + ( 5 − 6 ) 2 = ( − 4 ) 2 + ( − 1 ) 2 = 16 + 1 = 17
The length of BC
= ( 2 − 1 ) 2 + ( 1 − 5 ) 2 = 1 2 + ( − 4 ) 2 = 1 + 16 = 17 = \sqrt{(2 - 1)^2 + (1 - 5)^2}\\[1em] = \sqrt{1^2 + (-4)^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17} = ( 2 − 1 ) 2 + ( 1 − 5 ) 2 = 1 2 + ( − 4 ) 2 = 1 + 16 = 17
The length of CD
= ( 6 − 2 ) 2 + ( 2 − 1 ) 2 = 4 2 + 1 2 = 16 + 1 = 17 = \sqrt{(6 - 2)^2 + (2 - 1)^2}\\[1em] = \sqrt{4^2 + 1^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17} = ( 6 − 2 ) 2 + ( 2 − 1 ) 2 = 4 2 + 1 2 = 16 + 1 = 17
The length of DA
= ( 6 − 5 ) 2 + ( 2 − 6 ) 2 = 1 2 + 4 2 = 1 + 16 = 17 = \sqrt{(6 - 5)^2 + (2 - 6)^2}\\[1em] = \sqrt{1^2 + 4^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17} = ( 6 − 5 ) 2 + ( 2 − 6 ) 2 = 1 2 + 4 2 = 1 + 16 = 17
AB = BC = CD = DA = 17 \sqrt{17} 17
The length of diagonal AC =
= ( 2 − 5 ) 2 + ( 1 − 6 ) 2 = ( − 3 ) 2 + ( − 5 ) 2 = 9 + 25 = 34 = \sqrt{(2 - 5)^2 + (1 - 6)^2}\\[1em] = \sqrt{(-3)^2 + (-5)^2}\\[1em] = \sqrt{9 + 25}\\[1em] = \sqrt{34} = ( 2 − 5 ) 2 + ( 1 − 6 ) 2 = ( − 3 ) 2 + ( − 5 ) 2 = 9 + 25 = 34
The length of diagonal BD =
= ( 6 − 1 ) 2 + ( 2 − 5 ) 2 = ( − 5 ) 2 + ( − 3 ) 2 = 25 + 9 = 34 = \sqrt{(6 - 1)^2 + (2 - 5)^2}\\[1em] = \sqrt{(-5)^2 + (-3)^2}\\[1em] = \sqrt{25 + 9}\\[1em] = \sqrt{34} = ( 6 − 1 ) 2 + ( 2 − 5 ) 2 = ( − 5 ) 2 + ( − 3 ) 2 = 25 + 9 = 34
So, AC = BD
Since all sides and diagonals are equal, the points form a square.
Hence, the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.
