# Find the points on the $x$-axis which is equidistant from the points $(3,4)$ and $(1,-3)$.

Given:

Given points are $(3,4)$ and $(1,-3)$.

To do:

We have to find the point on x-axis which is equidistant from $(3,4)$ and $(1,-3)$.

Solution:

Let the coordinates of the two points be $A (3, 4)$ and $B (1, -3)$.

We know that,

The y co-ordinate of a point on x-axis is $0$.
Let the co-ordinates of the point which is equidistant from the points $A$ and $B$ be $C(x, 0)$.

This implies,

$AC = CB$

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.

Therefore,

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

$=\sqrt{(x-3)^{2}+16}$

$CB=\sqrt{(x-1)^{2}+(0+3)^{2}}$

$=\sqrt{(x-1)^{2}+9}$

$\Rightarrow \sqrt{(x-3)^{2}+16}=\sqrt{(x-1)^{2}+9}$

Squaring on both sides, we get,

$(x-3)^{2}+16=(x-1)^{2}+9$

$x^{2}-6 x+9+16=x^{2}-2 x+1+9$

$-6 x+2 x=10-25$

$-4x=-15$

$\Rightarrow x=\frac{15}{4}$

Therefore, the required point is $(\frac{15}{4}, 0)$.

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Updated on: 10-Oct-2022

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