Question

Find the point on X-axis which is equidistant from the points (2,6) and ($-$4,0)

Answer 1

Consider a point A(x, 0) on the X-axis,
which is equidistant from B(2, 6) and C(– 4, 0).
From these points, x₁ = 2, y₁ = 6, x₂ = – 4, y₂ = 0.

$$\begin{gathered} \mathrm{A}\mathrm{is}\mathrm{equidistan}\text{t}\mathrm{from}\mathrm{B}\mathrm{and}\mathrm{C} \\ \therefore \mathrm{AB}=\mathrm{AC} \\ \Rightarrow {\mathrm{AB}}^2={\mathrm{AC}}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(0-6\right)}^2={\left[x-\left(-4\right)\right]}^2+{\left(0-0\right)}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(-6\right)}^2={\left(x+4\right)}^2+{\left(0\right)}^2 \\ \Rightarrow x^2+4-4x+36=x^2+16+8x+0 \\ \Rightarrow 4-4x+36=16+8x+0 \\ \Rightarrow -4x+40=8x+16 \\ \Rightarrow -4x-8x=16-40 \\ \Rightarrow -12x=-24 \\ \Rightarrow x=2 \\ \mathrm{Therefore},\text{the}\mathrm{required}\mathrm{point}\mathrm{is}\left(2,0\right). \end{gathered}$$