Question

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

Answer 1

Answer:

Let the point equidistant from the points A(2,-5) and B(-2,9) be P(x,y).

Since the point P lies on x axis, y = 0

AP = BP as the point P is equidistant from A and B.

Applying distance formula

$\sqrt{(x - 2) {}^{2} + (0 - ( - 5)) {}^{2} } = \sqrt{( - 2 - x) {}^{2} + (9 - 0) {}^{2} }$

$\sqrt{x {}^{2} + 4 - 4x + 25} = \sqrt{4 + x {}^{2} + 4x + 81}$

Squaring both the sides

$x {}^{2} - 4x + 29 = x {}^{2} + 4x + 85$

$8x = - 56$

$x = - 7$

Therefore, the coordinates of the point equidistant from the points A and B are (-7,0)

Hope it helps...

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