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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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