CBSE Class 10 Math Question Paper 2016 | Q13

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

Video Explanation

Explanatory Answer | CBSE 2016 Maths Paper Q13

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}})
PA² = (a + b - x)² + (b - a - y)²
= a² + b² + x² + 2ab - 2bx - 2ax + b² + a² + y² - 2ab + 2ay - 2by
= 2a² +2b² + x² + y² - 2bx - 2ax + 2ay - 2by

Distance PB = \\sqrt{(a-b-x)^{2} + {{(a+b-y)}^2}})
PB² = (a - b - x)² + (a + b - y)²
= a² + b² + x² - 2ab + 2bx - 2ax + a² + b² + y² + 2ab - 2by - 2ay
= 2a² +2b² + x² + y² + 2bx - 2ax - 2by - 2ay

Because P is equidistant from A and B, PA = PB
∴ PA² = PB²

2a² + 2b² + x² + y² - 2bx - 2ax + 2ay - 2by = 2a² +2b² + x² + y² + 2bx - 2ax - 2by - 2ay
4ay = 4bx
So, ay = bx