# If point p (x,y) is equidistant if the point A(a+b,b-a) and B (a-b,a+b)prove that Bx =Ay

Here,

AP = √([x – (a – b)^2] + [y – (a + b)^2])

= √([x – a + b]^2 + [y – a -b]^2)

BP = √([x – (a + b)^2] + [y – (a – b)^2])

= √([x – a – b]^2 + [y – a + b]^2)

Since, AP = BP

√([x – a + b]^2 + [y – a -b]^2) = √([x – a – b]^2 + [y – a + b]^2)

On squaring both the sides, we get,

([x – a + b]^2 + [y – a -b]^2) = ([x – a – b]^2 + [y – a + b]^2)

x^2 + a^2 + b^2 – 2xa – 2ab + 2bx + y^2 + a^2 + b^2 – 2ya + 2ab – 2by = x^2 + a^2 + b^2 – 2xa + 2ab – 2bx + y^2 + a^2 + b^2 – 2yb – 2ab + 2ay

On solving, we get,

4bx = 4ay

Therefore, bx = ay