Question

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

Answer 1

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).

Answer 2

see the answer given above

hope you get it

thank you