Question

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

Answer 1

Answer:

(-7,0)

hope it help u dear....

Answer 2

Answer:

$\huge\underline\mathfrak\pink{❥Answer}$

Given points A(2,−5) and B(−2,9)Let the points be P(x,0).So, AP=PB and AP²=PB²

⇒(x−2)²+(0+5)²=(x+2)²+(0−9)

⇒x² +4−4x+25=x²+4+4x+81

=> x²+29-4x=x²+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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I hope this helps you.......