Find the coordinates of the point on x-axis which is equidistant from the points

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 $(-2,0)$.