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If $A (3, y)$ is equidistant from points $P (8, -3)$ and $Q (7, 6)$, find the value of $y$ and find the distance AQ.
Given:
$A (3, y)$ is equidistant from points $P (8, -3)$ and $Q (7, 6)$.
To do:
We have to find the value of $y$ and the distance AQ.
Solution:
Point $A (3, y)$ is equidistant from the points $P (8, -3)$ and $Q (7, 6)$.
This implies,
$AP = AQ$
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,
\( \sqrt{(3-8)^{2}+(y+3)^{2}}=\sqrt{(3-7)^{2}+(y-6)^{2}} \)
Squaring on both sides, we get,
\( (3-8)^{2}+(y+3)^{2}=(-4)^{2}+(y-6)^{2} \)
\( (-5)^{2}+y^{2}+6 y+9=16+y^{2}-12 y+36 \)
\( 25+y^{2}+6 y+9=16+y^{2}-12 y+36 \)
\( y^{2}+6 y-y^{2}+12 y=36-9-25+16 \)
\( 18 y=18 \)
\( \Rightarrow y=\frac{18}{18}=1 \)
\( \mathrm{AQ}=\sqrt{(3-8)^{2}+(1+3)^{2}} \)
\( =\sqrt{(-5)^{2}+(4)^{2}} \)
\( =\sqrt{25+16} \)
\( =\sqrt{41} \)
Therefore, the value of $y$ is $1$ and the distance $AQ$ is $\sqrt{41}$.
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