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Find the point on x axis be equidistant from p (2,-5) ,q(-2,9)​

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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).

hope it helps

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