Find a relation between x and y such that the point

Solution:

A(3,6) and B(–3, 4) are the given points. Point P (x, y) is equidistant from the points A and B.

$\Rightarrow \mathrm{PA}=\mathrm{PB}$

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

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

$\Rightarrow\left(x^{2}-6 x+9\right)+\left(y^{2}-12 y+36\right)$

$=\left(x^{2}+6 x+9\right)+\left(y^{2}-8 y+16\right)$

$\Rightarrow-6 x-12 y+45=6 x-8 y+25$

$\Rightarrow 12 x+4 y-20=0 \Rightarrow 3 x+y-5=0$