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

1.	Find the point on the x- axis which is equidistant from (2,-5) and (-2,9)
Answer» Given points\xa0A(2,−5)\xa0and\xa0B(−2,9)Let the points be\xa0P(x,0).So,\xa0AP=PB\xa0and\xa0AP2=PB2\xa0⇒(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=−7Hence, point on the\xa0x-axis which is equidistant from\xa0(2,−5)\xa0and\xa0(−2,9)\xa0is\xa0(−7,0).

Discussion