#### Question

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

#### Solution

Let P(x, y), Q(a + b, b – a) and R (a – b, a + b) be the given points. Then

PQ = PR (Given)

`=>sqrt({x-(a+b)}^2+{y-(b-a)}^2)=sqrt({x-(a-b)}^2+{y-(a+b)}^2`

`⇒ {x – (a + b)}^2 + {y – (b – a)}^2 = {x – (a – b)}^2 + {y – (a + b)}^2`

`⇒ x^2 – 2x (a + b) + (a + b)^2 + y^2 – 2y (b – a) + (b – a)^2 = x^2 + (a – b)^2 – 2x(a – b) + y^2 – 2 (a + b) + (a + b)^2`

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

⇒ ax + bx + by – ay = ax – bx + ay + by

⇒ 2bx = 2ay ⇒ bx = ay

Is there an error in this question or solution?

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