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Find a point on the x-axis which is equidistant from the points $(7, 6)$ and $(-3, 4)$.
Given:
Given points are $(7, 6)$ and $(-3, 4)$.
To do:
We have to find the point on x-axis which is equidistant from $(7, 6)$ and $(-3, 4)$.
Solution:
Let the co-ordinates of the two points be $A (7, 6)$ and $B (-3, 4)$.
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,
\( AC=\sqrt{(x-7)^2+(0-6)^2} \)
\( =\sqrt{(x-7)^{2}+36} \)
\( CB=\sqrt{(x+3)^{2}+(0-4)^{2}} \)
\( =\sqrt{(x+3)^{2}+16} \)
\( \Rightarrow \sqrt{(x-7)^{2}+36}=\sqrt{(x+3)^{2}+16} \)
Squaring on both sides, we get,
\( (x-7)^{2}+36=(x+3)^{2}+16 \)
\( x^{2}-14 x+49+36=x^{2}+6 x+9+16 \)
\( 6 x+14 x=85-25 \)
\( 20x=60 \)
\( \Rightarrow x=\frac{60}{20} \)
\( x=3 \)
Therefore, the required point is $(3,0)$.
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