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 $3x+y-5=0$.