Question

Find the point on X–axis which is equidistant from P(2,–5) and Q(–2,9).

Answer 1

Let the point on x-axis equidistant from P(2,–5) and Q(–2,9) be $\mathrm{A}\left(x,0\right)$.
$$\begin{gathered} \mathrm{AP}=\sqrt{{\left(x-2\right)}^2+{\left(0-\left(-5\right)\right)}^2}=\sqrt{{\left(x-2\right)}^2+25} \\ \mathrm{QA}=\sqrt{{\left(x-\left(-2\right)\right)}^2+{\left(0-9\right)}^2}=\sqrt{{\left(x+2\right)}^2+81} \\ \mathrm{AP}=\mathrm{QA} \\ \Rightarrow \sqrt{{\left(x-2\right)}^2+25}=\sqrt{{\left(x+2\right)}^2+81} \end{gathered}$$
Squaring both sides
$$\begin{gathered} {\left(x-2\right)}^2+25={\left(x+2\right)}^2+81 \\ \Rightarrow x^2+4-4x+25=x^2+4+4x+81 \\ \Rightarrow -8x=56 \\ \Rightarrow x=-7 \end{gathered}$$
Thus, the required point is $\left(-7,0\right)$.