Find the point on x-axis which is equidistant
Question:

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

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

Here we are to find out a point on the x-axis which is equidistant from both the points A(2,5) and B(2,3)

Let this point be denoted as C(x, y).

Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C

$A C=\sqrt{(-2-x)^{2}+(5-y)^{2}}$

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

$A C=\sqrt{(-2-x)^{2}+(5)^{2}}$

$B C=\sqrt{(2-x)^{2}+(-3-y)^{2}}$

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

$B C=\sqrt{(2-x)^{2}+(-3)^{2}}$

We know that both these distances are the same. So equating both these we get,

$A C=B C$

$\sqrt{(-2-x)^{2}+(5)^{2}}=\sqrt{(2-x)^{2}+(-3)^{2}}$

Squaring on both sides we have,

$(-2-x)^{2}+(5)^{2}=(2-x)^{2}+(-3)^{2}$

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

$8 x=-16$

$x=-2$

Hence the point on the $x$-axis which lies at equal distances from the mentioned points is $(-2,0)$.