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If the point $P (x, 3)$ is equidistant from the points $A (7, -1)$ and $B (6, 8)$, find the value of $x$ and find the distance AP.
Given:
The point $P (x, 3)$ is equidistant from the points $A (7, -1)$ and $B (6, 8)$.
To do:
We have to find the value of $x$ and the distance AP.
Solution:
Point $P (x, 3)$ is equidistant from the points $A (7, -1)$ and $B (6, 8)$.
This implies,
$PA = PB$
We know that,
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,
\( \sqrt{(x-7)^{2}+(3+1)^{2}}=\sqrt{(x-6)^{2}+(3-8)^{2}} \)
Squaring on both sides, we get,
\( (x-7)^{2}+(4)^{2}=(x-6)^{2}+(-5)^{2} \)
\( \Rightarrow x^{2}-14 x+49+16=x^{2}-12 x+36+25 \)
\( \Rightarrow x^{2}-14 x+65=x^{2}-12 x+61 \)
\( \Rightarrow x^{2}-14 x+12 x-x^{2}=61-65 \)
\( \Rightarrow-2 x=-4 \)
\( \Rightarrow x=\frac{-4}{-2}=2 \)
\( \Rightarrow x=2 \)
\( A P=\sqrt{(2-7)^{2}+(4)^{2}} \)
\( =\sqrt{(-5)^{2}+(4)^{2}} \)
\( =\sqrt{25+16} \)
\( =\sqrt{41} \)
Therefore, the value of $x$ is $2$ and the distance $AP$ is $\sqrt{41}$.
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