If P(x, y) is a point equidistant from the points A(6, −1) and B(2, 3), show that

The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)2 = (PB)2

$\Rightarrow(6-x)^{2}+(-1-y)^{2}=(2-x)^{2}+(3-y)^{2}$

$\Rightarrow x^{2}-12 x+36+y^{2}+2 y+1=x^{2}-4 x+4+y^{2}-6 y+9$

$\Rightarrow x^{2}+y^{2}-12 x+2 y+37=x^{2}+y^{2}-4 x-6 y+13$

$\Rightarrow x^{2}+y^{2}-12 x+2 y-x^{2}-y^{2}+4 x+6 y=13-37$

$\Rightarrow-8 x+8 y=-24$

$\Rightarrow-8(x-y)=-24$

$\Rightarrow x-y=\frac{-24}{-8}$

$\Rightarrow x-y=3$

Hence proved.