Question

Find the coordinates of the point on x-axis which is equidistant from the points (–2, 5) and (2, –3).

Answer 1

Let the point on the x - axis be (x, 0).
$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{A}\left(-2,5\right)\mathrm{and}\mathrm{B}\left(2,-3\right) \\ \mathrm{AX}=\mathrm{BX} \\ {\mathrm{AX}}^2={\mathrm{BX}}^2.....\left(1\right) \\ \mathrm{Using}\mathrm{distance}\mathrm{formula}: \\ d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ \mathrm{For}\mathrm{AX} \\ \mathrm{AX}=\sqrt{{\left(x-\left(-2\right)\right)}^2+{\left(0-5\right)}^2} \\ {\mathrm{AX}}^2={\left(x+2\right)}^2+{\left(-5\right)}^2 \\ {\mathrm{AX}}^2=x^2+4x+29.....\left(2\right) \end{gathered}$$
$$\begin{gathered} \mathrm{And} \\ \mathrm{BX}=\sqrt{{\left(x-2\right)}^2+{\left(0-\left(-3\right)\right)}^2} \\ {\mathrm{BX}}^2={\left(x-2\right)}^2+{\left(3\right)}^2 \\ {\mathrm{BX}}^2=x^2-4x+13.....\left(3\right) \\ \mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right) \\ {x}^2+4x+29=x^2-4x+13 \\ x=-2 \end{gathered}$$
Point on the x - axis is ($-$2, 0).