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Question
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.
Solution
It is given that A(3, y) is equidistant from points P(8, −3) and Q(7, 6).
∴ AP = AQ
Squaring on both sides, we get
\[25 + \left( y + 3 \right)^2 = 16 + \left( y - 6 \right)^2 \]
\[ \Rightarrow 25 + y^2 + 6y + 9 = 16 + y^2 - 12y + 36\]
\[ \Rightarrow 12y + 6y = 52 - 34\]
\[ \Rightarrow 18y = 18\]
\[ \Rightarrow y = 1\]
Thus, the value of y is 1.
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