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 x-axis. Therefore, its y-coordinate will be 0.

Let the point on 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

x² + 4-4x +25 = x²+4+4x+81

8x = 25-81

8x = -56

x =-7

Therefore, the point is (− 7, 0).