Question

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

Answer 1

Step-by-step explanation:

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