Question

17 If the point (x, y) is equidistant from the points (a+b,b-a) and
(a-bu+b), prove that bx =ay.
[CBSE 2011]​

Answer 1

Answer:

ANSWER

Let P(x,y), Q(a+b,b-a) and R(a-b,a+b) be the given points. Then,PQ=PR

⇒

{x−(a+b)}

2

+{y−(b−a)}

2

=

{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

−2y(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

REMARK-We know that a point which is equidistant from point P and Q lies on the

perpendicular bisector of PQ. Therefore, bx=ay is the equation of the perpendicular

bisector of PQ