Question

If the point P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k. [CBSE 2014]

Answer 1

It is given that P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5).

∴ AP = BP

$$\begin{gathered} \Rightarrow \sqrt{{\left[\left(k-1\right)-3\right]}^2+{\left(2-k\right)}^2}=\sqrt{{\left[\left(k-1\right)-k\right]}^2+{\left(2-5\right)}^2}\left(\mathrm{Distance}\mathrm{formula}\right) \\ \Rightarrow \sqrt{{\left(k-4\right)}^2+{\left(2-k\right)}^2}=\sqrt{{\left(-1\right)}^2+{\left(-3\right)}^2} \end{gathered}$$

Squaring on both sides, we get

$$\begin{gathered} k^2-8k+16+4-4k+k^2=10 \\ \Rightarrow 2k^2-12k+10=0 \\ \Rightarrow k^2-6k+5=0 \\ \Rightarrow \left(k-1\right)\left(k-5\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow k-1=0\mathrm{or}k-5=0 \\ \Rightarrow k=1\mathrm{or}k=5 \end{gathered}$$

Thus, the value of k is 1 or 5.