# Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3, 6)$ and $(-3, 4)$.

Given:

The points $(3, 6)$ and $(-3, 4)$ are equidistant from the point $(x, y)$.

To do:

We have to find a relation between $x$ and $y$.

Solution:

We know that,

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.

Therefore,

The distance between $(3,6)$ and $(x, y)$ $=\sqrt{(x-3)^{2}+(y-6)^{2}}$

The distance between $(-3,4)$ and $(x, y)$ $=\sqrt{(x+3)^{2}+(y-4)^{2}}$

The points $(3,6)$ and $(-3,4)$ are equidistant from $(x, y)$.

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

Squaring on both sides, we get,

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

$\Rightarrow x^{2}-6 x+9+y^{2}-12 y+36=x^{2}+6 x+9+y^{2}-8y+16$

$\Rightarrow x^{2}-6 x+y^{2}-12 y+36-x^{2}-6 x-y^{2}+8 y-16=0$

$\Rightarrow-12 x-4 y+20=0$

$\Rightarrow -4(3x+ y-5)=0$

$\Rightarrow 3x+y-5=0$

The required relation is $3x+y-5=0$.