If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
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