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Which point on x-axis is equidistant from $(5, 9)$ and $(-4, 6)$?
Given:
Given points are $(5, 9)$ and $(-4, 6)$.
To do:
We have to find the point on x-axis which is equidistant from $(5, 9)$ and $(-4, 6)$.
Solution:
Let the co-ordinates of the two points be $A (5, 9)$ and $B (-4, 6)$.
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,
\( AC=\sqrt{(x-5)^2+(0-9)^2} \)
\( =\sqrt{(x-5)^{2}+81} \)
\( CB=\sqrt{(x+4)^{2}+(0-6)^{2}} \)
\( =\sqrt{(x+4)^{2}+36} \)
\( \Rightarrow \sqrt{(x-5)^{2}+81}=\sqrt{(x+4)^{2}+36} \)
Squaring on both sides, we get,
\( (x-5)^{2}+81=(x+4)^{2}+36 \)
\( x^{2}-10 x+25+81=x^{2}+8 x+16+36 \)
\( -10 x-8 x=16+36-25-81 \)
\( -18x=52-106 \)
\( -18 x=-54 \)
\( \Rightarrow x=\frac{-54}{-18} \)
\( x=3 \)
Therefore, the required point is $(3,0)$.
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