Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9)
We have to find a point on x-axis.
Hence, its y-coordinate will be 0
Let the point on x-axis be (x, 0)
Distance between (x, 0) and (2, -5) = [(x- 2)2 + (0+ 5)2]1/2
= [(x- 2)2 + (5)2]1/2
Distance between (x, 0) and (-2, 9) = [(x + 2)2 + (0 + 9)2]1/2
= [(x + 2)2 + (9)2]1/2
By the given condition, these distances are equal in measure
[(x - 2)2 + (5)2]1/2 = [(x + 2)2 + (9)2]1/2
(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)