Question

find the point x axis with is equal distance from (2, -5), (-2, 9)​

Answer 1

Answer:

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).Read more on Sarthaks.com - https://www.sarthaks.com/2428/find-the-point-on-the-x-axis-which-is-equidistant-from-2-5-and-2-9