Find a point on the x-axis which is equidistant


Find a point on the x-axis which is equidistant from the points (7, 6) and (−3, 4).


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


Here we are to find out a point on the x−axis which is equidistant from both the points A(7,6) and B(3,4).

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

$A C=\sqrt{(7-x)^{2}+(6-y)^{2}}$


$A C=\sqrt{(7-x)^{2}+(6)^{2}}$

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


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

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

$A C=B C$


Squaring on both sides we have,


$49+x^{2}-14 x+36=9+x^{2}+6 x+16$

$20 x=60$


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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now