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Find the point on x-axis which is equidistant from the points $(-2, 5)$ and $(2, -3)$.
Given:
Given points are $(-2, 5)$ and $(2, -3)$.
To do:
We have to find the point on x-axis which is equidistant from $(-2, 5)$ and $(2, -3)$.
Solution:
Let the co-ordinates of the two points be $A (-2, 5)$ and $B (2, -3)$.
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,
\( AC=\sqrt{(x+2)^2+(0-5)^2} \)
\( =\sqrt{(x+2)^{2}+25} \)
\( CB=\sqrt{(x-2)^{2}+(0+3)^{2}} \)
\( =\sqrt{(x-2)^{2}+9} \)
\( \Rightarrow \sqrt{(x+2)^{2}+25}=\sqrt{(x-2)^{2}+9} \)
Squaring on both sides, we get,
\( (x+2)^{2}+25=(x-2)^{2}+9 \)
\( x^{2}+4 x+4+25=x^{2}-4 x+4+9 \)
\( 4 x+4 x=9-25 \)
\( 8x=-16 \)
\( \Rightarrow x=\frac{-16}{8} \)
\( x=-2 \)
Therefore, the required point is $(-2,0)$.
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