Question:
Find the coordinates of the point on x-axis which is equidistant from the points (–2, 5) and (2, –3).
Solution:
Let the point on the x - axis be (x, 0).
We have $\mathrm{A}(-2,5)$ and $\mathrm{B}(2,-3)$
$\mathrm{AX}=\mathrm{BX}$
$\mathrm{AX}^{2}=\mathrm{BX}^{2} \quad \ldots \ldots(1)$
Using distance formula :
$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
For AX
$\mathrm{AX}=\sqrt{(x-(-2))^{2}+(0-5)^{2}}$
$\mathrm{AX}^{2}=(x+2)^{2}+(-5)^{2}$
$\mathrm{AX}^{2}=x^{2}+4 x+29 \quad \ldots \ldots(2)$
And
$\mathrm{BX}=\sqrt{(x-2)^{2}+(0-(-3))^{2}}$
$\mathrm{BX}^{2}=(x-2)^{2}+(3)^{2}$
$\mathrm{BX}^{2}=x^{2}-4 x+13 \quad \ldots . .(3)$
From $(1),(2)$ and $(3)$
$x^{2}+4 x+29=x^{2}-4 x+13$
$x=-2$
Point on the $x$ - axis is $(-2,0)$.