Question

Find the value of k , if the point P (0,2) is equidistant from (3, k ) and (k, 5) .

Answer 1

It is given that P(0, 2) is equidistant from A(3, k) and B(k, 5).
So, AP = PB
$$\begin{gathered} \Rightarrow A{P}^2=P{B}^2 \\ \Rightarrow {\left(3-0\right)}^2+{\left(k-2\right)}^2={\left(k-0\right)}^2+{\left(5-2\right)}^2\left[Usedis\tan ceformula\right] \\ \Rightarrow 9+k^2-4k+4=k^2+9 \\ \Rightarrow -4k+4=0 \\ \Rightarrow k=1 \end{gathered}$$
Therefore, the value of k is 1.