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If the point P(x-y) is equidistant from the points A(a+b,b-a) and B(a-b,a+b) then prove that bx = ay
Distace between the points (x, y) and (a+b, b-a) & (a-b, a+b) is equal ⇒ √{[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)⇒ -2ax - 2bx - 2by + 2ay = - 2ax + 2bx - 2ay - 2by⇒ ay - bx = bx - ay⇒ 2ay = 2bx⇒ bx = ay
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