1. Find the coordinates of the point equidistant from the points A(1,

1.	1. Find the coordinates of the point equidistant from the points A(1, 2), B (3.-4) and C(5.-6).(a) (2,3)(d) (1,3)SIA
Answer» GIVEN , The COORDINATES of the point equidistant from the POINTS A(1, 2) , B(3.-4) and C(5.-6) Let , the coordinate of point P(x , y) We know that , the distance between two points is given by $\boxed{ \sf{Distance = \sqrt{ {( x_{2} - x_{1} )}^{2} + {(y_{2} - y_{1} )}^{2} } }}$ ACCORDING to the question , AB = BP $\tt \sqrt{ {(x - 1)}^{2} + {(y - 2)}^{2} } = \sqrt{ {(x - 3)}^{2} + {(y + 4)}^{2}}$ $\tt {(x)}^{2} + 1 - 2x + {(y)}^{2} + 4 - 4y = {(x)}^{2} + 9 - 6y + {(y)}^{2} + 16 + 8y$ $\tt 5 - 2x - 4y = 25 - 6x + 8y$ $\tt 4x = 20 + 12y$ $\tt 4x - 12y = 20$ $\tt x - 3y = 5 - - - (i)$ And BP = CP $\tt \sqrt{ {(x - 3)}^{2} + {(y + 4)}^{2} } = \sqrt{ {(x - 5)}^{2} + {(y + 6)}^{2} }$ $\tt {(x)}^{2} + 9 - 6x + {(y)}^{2} + 16 + 8y = {(x)}^{2} + 25 - 10x + {(y)}^{2} + 36 + 12y$ $\tt 25 - 6x + 8y = 61 - 10x + 12y$ $\tt 4x - 4y = 36$ $\tt x - y = 9 - - - (ii)$ Subtract eq (i) from eq (ii) , we get x - y - (x - 3y) = 9 - 5 3y - y = 4 2y = 4 y = 2 Put y = 2 in eq (ii) , we get x - 2 = 9 x = 11 ★ The coordinates of the point equidistant from the points A , B and C is (11,2)

1. Find the coordinates of the point equidistant from the points A(1, 2), B (3.-4) and C(5.-6).(a) (2,3)(d) (1,3)SIA

Answer»

GIVEN ,

The COORDINATES of the point equidistant from the POINTS A(1, 2) , B(3.-4) and C(5.-6)

Let , the coordinate of point P(x , y)

We know that , the distance between two points is given by

$\boxed{ \sf{Distance = \sqrt{ {( x_{2} - x_{1} )}^{2} + {(y_{2} - y_{1} )}^{2} } }}$

ACCORDING to the question ,

AB = BP

$\tt \sqrt{ {(x - 1)}^{2} + {(y - 2)}^{2} } = \sqrt{ {(x - 3)}^{2} + {(y + 4)}^{2}}$

$\tt {(x)}^{2} + 1 - 2x + {(y)}^{2} + 4 - 4y = {(x)}^{2} + 9 - 6y + {(y)}^{2} + 16 + 8y$

$\tt 5 - 2x - 4y = 25 - 6x + 8y$

$\tt 4x = 20 + 12y$

$\tt 4x - 12y = 20$

$\tt x - 3y = 5 - - - (i)$

And

BP = CP

$\tt \sqrt{ {(x - 3)}^{2} + {(y + 4)}^{2} } = \sqrt{ {(x - 5)}^{2} + {(y + 6)}^{2} }$

$\tt {(x)}^{2} + 9 - 6x + {(y)}^{2} + 16 + 8y = {(x)}^{2} + 25 - 10x + {(y)}^{2} + 36 + 12y$

$\tt 25 - 6x + 8y = 61 - 10x + 12y$

$\tt 4x - 4y = 36$

$\tt x - y = 9 - - - (ii)$

Subtract eq (i) from eq (ii) , we get

x - y - (x - 3y) = 9 - 5

3y - y = 4

2y = 4

y = 2

Put y = 2 in eq (ii) , we get

x - 2 = 9

x = 11

★ The coordinates of the point equidistant from the points A , B and C is (11,2)