Question

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

Answer 1

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)2

⇒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).

Step-by-step explanation:

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