Question

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

Answer 1

HEYA..

HERE IS YOUR ANSWER..

   Let the point of x-axis be P(x, 0)

Given A(2, -5) and B(-2, 9) are equidistant from P

That is PA = PB

Hence PA2 = PB2  → (1)

Distance between two points is √[(x2 - x1)2 + (y2 - y1)2]

PA = √[(2 - x)2 + (-5 - 0)2]

PA2 = 4 - 4x +x2 + 25 = x2 - 4x + 29

Similarly, PB2 = x2 + 4x + 85

Equation (1) becomes

x2 - 4x + 29 = x2 + 4x + 85

- 8x = 56

x = -7

Hence the point on x-axis is (-7, 0)

IT MAY HELP YOU..

Answer 2

Hey mate ^_^

Check this attachment.

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