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).