# Find the point on x-axis which is equidistant from the points $(-2, 5)$ and $(2, -3)$.

Given:

Given points are $(-2, 5)$ and $(2, -3)$.

To do:

We have to find the point on x-axis which is equidistant from $(-2, 5)$ and $(2, -3)$.

Solution:

Let the co-ordinates of the two points be $A (-2, 5)$ and $B (2, -3)$.

We know that,

The y co-ordinate of a point on x-axis is $0$.
Let the co-ordinates of the point which is equidistant from the points $A$ and $B$ be $C(x, 0)$.

This implies,

$AC = CB$

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.

Therefore,

$AC=\sqrt{(x+2)^2+(0-5)^2}$

$=\sqrt{(x+2)^{2}+25}$

$CB=\sqrt{(x-2)^{2}+(0+3)^{2}}$

$=\sqrt{(x-2)^{2}+9}$

$\Rightarrow \sqrt{(x+2)^{2}+25}=\sqrt{(x-2)^{2}+9}$

Squaring on both sides, we get,

$(x+2)^{2}+25=(x-2)^{2}+9$

$x^{2}+4 x+4+25=x^{2}-4 x+4+9$

$4 x+4 x=9-25$

$8x=-16$

$\Rightarrow x=\frac{-16}{8}$

$x=-2$

Therefore, the required point is $(-2,0)$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

27 Views