Question

Find the point on the y-axis which is equidistant from the points A(6, 5) and B(− 4, 3).

Answer 1

Let P (0, y) be a point on the y-axis. Then as per the question, we have
$$\begin{gathered} AP=BP \\ \Rightarrow \sqrt{{\left(0-6\right)}^2+{\left(y-5\right)}^2}=\sqrt{{\left(0+4\right)}^2+{\left(y-3\right)}^2} \\ \Rightarrow \sqrt{{\left(6\right)}^2+{\left(y-5\right)}^2}=\sqrt{{\left(4\right)}^2+{\left(y-3\right)}^2} \\ \Rightarrow {\left(6\right)}^2+{\left(y-5\right)}^2={\left(4\right)}^2+{\left(y-3\right)}^2\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow 36+y^2-10y+25=16+y^2-6y+9 \\ \Rightarrow 4y=36 \\ \Rightarrow y=9 \end{gathered}$$
Hence, the point on the y-axis is (0, 9).