Question

Find the value of k, if the point P (2, 4) is equidistant from the points A (5, k) and B (k, 7).

Answer 1

P (2, 4) is equidistant from the points A (5, k) and B (k, 7)
$$\begin{gathered} \mathrm{AP}=\mathrm{BP} \\ \Rightarrow \sqrt{{\left(2-5\right)}^2+{\left(4-k\right)}^2}=\sqrt{{\left(2-k\right)}^2+{\left(4-7\right)}^2} \\ \Rightarrow {\left(2-5\right)}^2+{\left(4-k\right)}^2={\left(k-2\right)}^2+{\left(4-7\right)}^2 \\ \Rightarrow {\left(4-k\right)}^2={\left(2-k\right)}^2 \\ \Rightarrow 16-8k+k^2=4-4k+k^2 \\ \Rightarrow 4k=12 \\ \Rightarrow k=3 \end{gathered}$$
Hence, the value of k is 3