If $P(x, y)$ is equidistant from the points $A(a+b, b-a)$ and $B(a-b, a+b),$ then prove that $bx = ay.$

(N/A) Given that point $P(x, y)$ is equidistant from points $A(a+b, b-a)$ and $B(a-b, a+b).$
By the distance formula,$PA = PB.$
Squaring both sides,$PA^2 = PB^2.$
Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2,$
$(x - (a+b))^2 + (y - (b-a))^2 = (x - (a-b))^2 + (y - (a+b))^2.$
Expanding both sides:
$x^2 - 2x(a+b) + (a+b)^2 + y^2 - 2y(b-a) + (b-a)^2 = x^2 - 2x(a-b) + (a-b)^2 + y^2 - 2y(a+b) + (a+b)^2.$
Canceling $x^2, y^2, (a+b)^2$ from both sides:
$-2x(a+b) - 2y(b-a) + (b-a)^2 = -2x(a-b) - 2y(a+b) + (a-b)^2.$
Since $(b-a)^2 = (a-b)^2,$ these terms also cancel out:
$-2ax - 2bx - 2by + 2ay = -2ax + 2bx - 2ay - 2by.$
Canceling $-2ax$ and $-2by$ from both sides:
$-2bx + 2ay = 2bx - 2ay.$
Rearranging the terms:
$2ay + 2ay = 2bx + 2bx.$
$4ay = 4bx.$
Dividing by $4,$ we get $bx = ay.$