Question

Find the point on the x-axis which is equidistant from the points (2, −5) and (−2, 9).

Answer 1

Let (x, 0) be the point on the x-axis. Then as per the question, we have
$$\begin{gathered} \sqrt{{\left(x-2\right)}^2+{\left(0+5\right)}^2}=\sqrt{{\left(x+2\right)}^2+{\left(0-9\right)}^2} \\ \Rightarrow \sqrt{{\left(x-2\right)}^2+{\left(5\right)}^2}=\sqrt{{\left(x+2\right)}^2+{\left(-9\right)}^2} \\ \Rightarrow {\left(x-2\right)}^2+{\left(5\right)}^2={\left(x+2\right)}^2+{\left(-9\right)}^2\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow x^2-4x+4+25=x^2+4x+4+81 \end{gathered}$$
$$\begin{gathered} \Rightarrow 8x=25-81 \\ \Rightarrow x=-\frac{56}{8}=-7 \end{gathered}$$
Hence, the point on the x-axis is (− 7, 0).