Question

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

Answer 1

Let the given points be A(−2, 5) and B(−2, 9) and the required point be P(x,0). Then, AP = PB.
That is, (AP)² = (PB)²
$$\begin{gathered} \Rightarrow {\left\{x-\left(-2\right)\right\}}^2+{\left(0-5\right)}^2={\left(-2-x\right)}^2+{\left(9-0\right)}^2 \\ \Rightarrow {\left(x+2\right)}^2+{\left(0-5\right)}^2={\left(-2-x\right)}^2+{\left(9\right)}^2 \\ \Rightarrow {\left(x+2\right)}^2+{\left(-5\right)}^2={\left(-2-x\right)}^2+{\left(9\right)}^2 \\ \Rightarrow {\left(x+2\right)}^2+25={\left(-2-x\right)}^2+81 \\ \Rightarrow x^2+4x+4+25=x^2+4x+4+81 \\ \Rightarrow x^2+4x+29=x^2+4x+85 \\ \Rightarrow x^2+4x-x^2-4x=85-29 \\ \Rightarrow 0=56 \end{gathered}$$

Disclaimer : As it is never possible that 0 will become equal to 56
Therefore, we can say that there is no point on the x-axis which is equidistance from A(-2,5) and B(-2,9)