# If the point (x,

Let point P(x,y) , A(a + b,a – b) , B(a – b,a + b)

Then AP = BP

AP = √{((a + b) – x)2 + ((a – b) – y)2}

= √{(a + b)2 + x2 – 2(a + b)x + (a – b)2 + y2 – 2(a – b)y}

= √(a2 + b2 + 2ab + x2 – 2(a + b)x + b2 + a2 – 2ab + y2 – 2(a – b)y)

BP = √{((a – b) – x)2 + ((a + b) – y)2}

= √{(a – b)2 + x2 – 2(a – b)x + (a + b)2 + y2 – 2(a + b)y}

= √(a2 + b2 – 2ab + x2 – 2(a – b)x + b2 + a2 + 2ab + y2 – 2(a + b)y)

Squaring and Equating both we get

a2 + b2 + 2ab + x2 – 2(a + b)x + b2 + a2 – 2ab + y2 – 2(a – b)y = a2 + b2 – 2ab + x2 – 2(a – b)x + b2 + a2 + 2ab + y2 – 2(a + b)y

– 2(a + b)x – 2(a – b)y = – 2(a – b)x – 2(a + b)y

ax + bx + ay – by = ax – bx + ay + by

Hence

bx = ay

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