Class 10

Math

All topics

Coordinate Geometry

Find the point of the x-axis which is equidistant from $(2,−5)$ and $(−2,9)$.

Let point on $X−$ axis be $P(x,0)$ to be equidistant from $A(2,−5)$ and $B(−2,9)$

$⇒PA=PB$

$⇒(x−2)_{2}+(0+5)_{2} =(x+2)_{2}+(0−9)_{2} $

$⇒PA=PB$

$⇒(x−2)_{2}+(0+5)_{2} =(x+2)_{2}+(0−9)_{2} $

$⇒(x−2)_{2}+25=(x+2)_{2}+81$

$⇒−56=8x$

$⇒x=−7$

So, point equidistant from $(2,−5)$ and lying on $x$axis is $(−7,0)$