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 $(-2,0)$.