If a point P(x, y) is equidistant from the points A

Given: Point P(x, y) is equidistant from points A(6, -1) and B(2, 3) i.e., distance of P from A = distance of P from B

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

Squaring both sides

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

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

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

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

$\Rightarrow x-y=3$

Therefore, x – y = 3 is the required relation.