Question

Hello guys .
Find the point on the x-axis which is equidistant for from (2,-5) and (-2,9)
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Answer 1

Answer:

Given points A(2,−5) and B(−2,9)

Let the points be P(x,0).

So, AP=PB and AP^2  =PB^2

 (x−2)^2  +(0+5)^2  =(x+2)^2  +(0−9)^2  

x^2  +4−4x+25=x^2  +4+4x+81

x^2  +29−4x=x^2  +85+4x

−4x−4x=85−29

−8x=56

x=−7

Hence, point on the x-axis which is equidistant from (2,−5) and (−2,9) is (−7,0).

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