Which point on x-axis is equidistant from (5, 9) and (−4, 6) ?
Question:

Which point on x-axis is equidistant from (5, 9) and (−4, 6) ?

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 (59) and (46)

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

$A C=\sqrt{(5-x)^{2}+(9-y)^{2}}$

$=\sqrt{(5-x)^{2}+(9-0)^{2}}$

$A C=\sqrt{(5-x)^{2}+(9)^{2}}$

$B C=\sqrt{(-4-x)^{2}+(6-y)^{2}}$

$=\sqrt{(4+x)^{2}+(6-0)^{2}}$

$B C=\sqrt{(4+x)^{2}+(6)^{2}}$

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

$A C=B C$

$\sqrt{(5-x)^{2}+(9)^{2}}=\sqrt{(4+x)^{2}+(6)^{2}}$

Squaring on both sides we have,

$(5-x)^{2}+(9)^{2}=(4+x)^{2}+(6)^{2}$

$25+x^{2}-10 x+81=16+x^{2}+8 x+36$

$18 x=54$

$x=3$

Hence the point on the $x$-axis which lies at equal distances from the mentioned points is $(3,0)$.