The given question is a medium difficulty 3-mark question from the chapter Coordinate Geometry. Concept: Distance formula when coordinates of two points are given. This question appeared in Section C of the 2016 CBSE class 10 board paper.

Question 13: If the point P(x, y) is equidistant from the points A(a + b, b - a) and P(a - b, a + b). Prove that bx = ay

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P(x, y) is equidistant from A(a + b, b - a) and B(a - b, a + b).

So, PA = PB

Distance PA = \\sqrt{(a+b-x)^{2} + {{(b - a - y)}^2}})

PA^{2} = (a + b - x)^{2} + (b - a - y)^{2}

= a^{2} + b^{2} + x^{2} + 2ab - 2bx - 2ax + b^{2} + a^{2} + y^{2} - 2ab + 2ay - 2by

= 2a^{2} +2b^{2} + x^{2} + y^{2} - 2bx - 2ax + 2ay - 2by

Distance PB = \\sqrt{(a-b-x)^{2} + {{(a+b-y)}^2}})

PB^{2} = (a - b - x)^{2} + (a + b - y)^{2}

= a^{2} + b^{2} + x^{2} - 2ab + 2bx - 2ax + a^{2} + b^{2} + y^{2} + 2ab - 2by - 2ay

= 2a^{2} +2b^{2} + x^{2} + y^{2} + 2bx - 2ax - 2by - 2ay

Because P is equidistant from A and B, PA = PB

∴ PA^{2} = PB^{2}

2a^{2} + 2b^{2} + x^{2} + y^{2} - 2bx - 2ax + 2ay - 2by = 2a^{2} +2b^{2} + x^{2} + y^{2} + 2bx - 2ax - 2by - 2ay

4ay = 4bx

So, ay = bx

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