Question

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

Answer 1

We have to find a point on the x-axis. Therefore, its y-coordinate will be 0. Let the point on the x-axis be (x,0).
Distance between (x,0) and (2,-5)=\(\sqrt{(x-2)^2+(0-(-5))^2}=\sqrt{(x-2)^2+(5)^2}\)
Distance between (x,0) and (-2,9)=\(\sqrt{(x-(-2))^2+(0-(-9))^2}=\sqrt{(x+2)^2+(9)^2}\)
By the given condition, these distances are equal in measure.
\(\sqrt{(x-2)^2+(5)^2}=\sqrt{(x+2)^2+(9)^2}\)
\((x-2)^2+25=(x+2)^2+81\)
\(x^2+4-4x+25=x^2+4+4x+81\)
\(8x=25-81\)
\(8x=-56\)
\(x=-7\)
Therefore, the point is (− 7, 0).