# Alg6.7

Find the relation between x and y,if the point (x,y) is equidistant from $\displaystyle (7,-6),(-3,4)\,$

The distance between (x,y) and the first point is

$\displaystyle \sqrt{(x-7)^2+(y+6)^2}\,$

$\displaystyle \sqrt{x^2+49-14x+y^2+36+12y}\,$

$\displaystyle \sqrt{x^2+y^2-14x+12y+85}\,$

Let this expression be 1.

The distance between (x,y) and the second point is

$\displaystyle \sqrt{(x+3)^2+(y-4)^2}\,$

$\displaystyle \sqrt{x^2+9+6x+y^2+16-8y}\,$

$\displaystyle \sqrt{x^2+y^2+6x-8y+25}\,$

Let this expression be 2.

As per the given condition,equating both the expressions,

$\displaystyle \sqrt{x^2+y^2-14x-12y+85}=\sqrt{x^2+y^2+6x-8y+25}\,$

Squaring on both sides,we get

$\displaystyle x^2+y^2-14x+12y+22=x^2+y^2+6x-8y+25\,$

cancelling the terms which are common on both sides,

$\displaystyle -14x-6x+12y+8y=-60\,$

$\displaystyle -20x+20y=-60\,$

$\displaystyle x-y=3\,$