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

Since the point P(x,y) is equidistant from the points A(a+b,b-a)≡(x1,y1) and B(a-b,a+b)≡(x2,y2).
Using distance formula,
Distance Formula = √(x−x1)²+(y−y1)²
PA = √(x−(a+b))²+(y+(b−a))²
PB = √(x−(a−b))²+(y+(a+b))²
Therefore, PA=PB
√(x−(a+b))²+(y+(b−a))² = √(x−(a−b))²+(y+(a+b))²
Squaring on both the sides,
(x−(a+b))²+(y+(b−a))² = (x−(a−b))²+(y+(a+b))²
x²+(a+b)²−2x(a+b)+y²+(b−a)²−2y(b−a) = x²+(a−b)²−2x(a−b)+y²+(a+b)²−2y(a+b)
x²+(a+b)²−2x(a+b)+y²+(b−a)²−2y(b−a) = x²+(a−b)²−2x(a−b)+y²+(a+b)²−2y(a+b)
−2x(a+b)−2y(b−a) = −2x(a−b)−2y(a+b)
−ax−bx−by+ay = −ax+bx−ay−by
−bx+ay = bx−ay
2ay = 2bx
bx = ay
Hence proved.