Question

If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k. Also find the length of AP.

Answer 1

As per the question, we have
$$\begin{gathered} AP=BP \\ \Rightarrow \sqrt{{\left(2+2\right)}^2+{\left(2-k\right)}^2}=\sqrt{{\left(2+2k\right)}^2+{\left(2+3\right)}^2} \\ \Rightarrow \sqrt{{\left(4\right)}^2+{\left(2-k\right)}^2}=\sqrt{{\left(2+2k\right)}^2+{\left(5\right)}^2} \\ \Rightarrow 16+4+k^2-4k=4+4k^2+8k+25\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow k^2+4k+3=0 \\ \Rightarrow \left(k+1\right)\left(k+3\right)=0 \\ \Rightarrow k=-3,-1 \end{gathered}$$
Now for $k=-1$
$$\begin{gathered} AP=\sqrt{{\left(2+2\right)}^2+{\left(2-k\right)}^2} \\ =\sqrt{{\left(4\right)}^2+{\left(2+1\right)}^2} \\ =\sqrt{16+9}=5\mathrm{units} \end{gathered}$$
For $k=-3$
$$\begin{gathered} AP=\sqrt{{\left(2+2\right)}^2+{\left(2-k\right)}^2} \\ =\sqrt{{\left(4\right)}^2+{\left(2+3\right)}^2} \\ =\sqrt{16+25}=\sqrt{41}\mathrm{units} \end{gathered}$$
Hence, $k=-1,-3;AP=5\mathrm{units}\mathrm{for}k=-1\mathrm{and}AP=\sqrt{41}\mathrm{units}\mathrm{for}k=-3$.