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. [CBSE 2014]

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

$$\begin{gathered} 25+{\left(y+3\right)}^2=16+{\left(y-6\right)}^2 \\ \Rightarrow 25+y^2+6y+9=16+y^2-12y+36 \\ \Rightarrow 12y+6y=52-34 \\ \Rightarrow 18y=18 \\ \Rightarrow y=1 \end{gathered}$$

Thus, the value of y is 1.

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