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)


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