Question

If the points $(x,y)$ be equidistant from the points $(6,-1)$ and $(2,3)$ prove that $x-y=3$

Answer 1

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

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

$37-12x+2y=-4x-6y+13$

$24=8x-8y$

$x-y=3$