# CBSE Class 10 Math Question Paper 2016 | Q13

###### CBSE Maths Board Paper Solution | 3 Mark Question | Coordinate Geometry

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

## NCERT Solution to Class 10 Maths

### 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

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