Question

If a point P(x, y) is equidistant from the points A (6,-1) and B(2,3)find the relation between X and Y.

Answer 1

Distance between two points
$= \sqrt{(x2 -x1) {}^{2} + (y 2 - y1) {}^{2} }$


AP=
$\sqrt{(x - 6) {}^{2} + (y - ( - 1) {}^{2} }$


$= \sqrt{(x {}^{2} + 36 - 12x) + (y + 1) {}^{2} }$


$= \sqrt{ {x}^{2} + 36 - 12x + {y}^{2} + 1 + 2y}$


$= \sqrt{ {x}^{2} + {y}^{2} - 12x + 2y + 37 }$

AP^2
$= ( \sqrt{ {x}^{2} + y {}^{2} - 12x + 2y + 37 } ){}^{2}$


$= {x}^{2} + {y}^{2} - 12x + 2y + 37$


BP
$= \sqrt{(x - 2) {}^{2} + (y - 3) {}^{2} }$


$= \sqrt{ {x}^{2} + 4 - 4x + {y}^{2} + 9 - 6y}$


$= \sqrt{ {x}^{2} + {y}^{2} - 4x - 6y + 13 }$


BP^2
$= (\sqrt{x {}^{2} + y {}^{2} - 4x - 6y + 13} ) {}^{2}$


$= {x}^{2} + {y}^{2} - 4x - 6y + 13$


AP^2 = BP^2

x^2 + y^2 -12 x +2 y +37 = x^2 + y^2 -4x -6y +13

-12x+2y +37 = -4x -6y +13

2y +6y = -4x +12 x +13-37

8y = 8x - 24

dividing the equation by 8 .

y=x - 3

3 = x-y

This is the required relationship.