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. [CBSE 2014]

Answer 1

It is given that P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3).

∴ AP = BP

$\Rightarrow \sqrt{{\left(-2-2\right)}^2+{\left(k-2\right)}^2}=\sqrt{{\left(-2k-2\right)}^2+{\left(-3-2\right)}^2}$ (Distance formula)

Squaring on both sides, we get

$$\begin{gathered} 16+k^2-4k+4=4k^2+8k+4+25 \\ \Rightarrow 3k^2+12k+9=0 \\ \Rightarrow k^2+4k+3=0 \\ \Rightarrow \left(k+3\right)\left(k+1\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow k+3=0\mathrm{or}k+1=0 \\ \Rightarrow k=-3\mathrm{or}k=-1 \end{gathered}$$

Thus, the value of k is −1 or −3.

When k = −1,

$\mathrm{AP}=\sqrt{{\left(-2-2\right)}^2+{\left(-1-2\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}=\sqrt{16+9}=\sqrt{25}=5$ units

When k = −3,

$\mathrm{AP}=\sqrt{{\left(-2-2\right)}^2+{\left(-3-2\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2}=\sqrt{16+25}=\sqrt{41}$ units