Question

Finda point on the x axis which is equidistant from the point A(7,6) and B (-3,4)

Answer 1

point on x axis be P (x,0) since it's on x axis
ATQ
AP=BP i.e equidistant,
now,

√(7-x)²+(6-0)² = √(-3-x)²+ (4-0)²
squaring both sides we get ,
(7-x)²+(6)²=(-3-x)²+(4)². (roots are removed)
(7)²+(x)²+2(7)(x)+(6)²=(-3)²+(-x)²+2(-3)(-x)+(4)²
49+x²+14x+36=9+x²+6x+16
after solving we get,
49+36-9-16+14x-6x-x²+x²=0
60+8x=0
8x=-60
x=-60/8
x=7.5


plz mark brainliest....