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. 

Answer 1

It is given that A(3, y) is equidistant from points P(8, −3) and Q(7, 6).
∴ AP = AQ

\[\Rightarrow \sqrt{\left( 3 - 8 \right)^2 + \left[ y - \left( - 3 \right) \right]^2} = \sqrt{\left( 3 - 7 \right)^2 + \left( y - 6 \right)^2}\]           (Distance formula)

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.

\[\therefore AQ = \sqrt{\left( 3 - 7 \right)^2 + \left( 1 - 6 \right)^2} = \sqrt{\left( - 4 \right)^2 + \left( - 5 \right)^2} = \sqrt{16 + 25} = \sqrt{41}\]  units