Write the coordinates of a point on X-axis which is equidistant

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(-3,4) and B(2,5).

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’

$A C=\sqrt{(-3-x)^{2}+(4-y)^{2}}$

$=\sqrt{(-3-x)^{2}+(4-0)^{2}}$

$\mathrm{AC}=\sqrt{(-3-x)^{2}+(4)^{2}}$

$\mathrm{BC}=\sqrt{(2-x)^{2}+(5-y)^{2}}$

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

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

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

$\mathrm{AC}=\mathrm{BC}$

$\sqrt{(-3-x)^{2}+(4)^{2}}=\sqrt{(2-x)^{2}+(5)^{2}}$

Squaring on both sides we have,

$(-3-x)^{2}+(4)^{2}=(2-x)^{2}+(5)^{2}$

$9+x^{2}+6 x+16=4+x^{2}-4 x+25$

$10 x=4$

$x=\frac{2}{5}$

Hence the point on the $x$-axis which lies at equal distances from the mentioned points is $\left(\frac{2}{5}, 0\right)$