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

The given points are P(k − 1, 2), A(3, k) and B(k, 5).
$$\begin{gathered} \because AP=BP \\ \therefore A{P}^2=B{P}^2 \\ \Rightarrow {\left(k-1-3\right)}^2+{\left(2-k\right)}^2={\left(k-1-k\right)}^2+{\left(2-5\right)}^2 \\ \Rightarrow {\left(k-4\right)}^2+{\left(2-k\right)}^2={\left(-1\right)}^2+{\left(-3\right)}^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow k^2-8y+16+4+k^2-4k=1+9 \\ \Rightarrow k^2-6y+5=0 \\ \Rightarrow \left(k-1\right)\left(k-5\right)=0 \\ \Rightarrow k=1\mathrm{or}k=5 \end{gathered}$$
Hence, k = 1 or k = 5.