The coordinates of a point on x-axis which lies on the perpendicular

TO FIND: The coordinates of a point on x axis which lies on perpendicular bisector of line segment joining points (7, 6) and (−3, 4).

Let P(x, y) be any point on the perpendicular bisector of AB. Then,

PA=PB

$\sqrt{(x-7)^{2}+(y-6)^{2}}=\sqrt{(x-(-3))^{2}+(y-4)^{2}}$

$(x-7)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2}$

$x^{2}-14 x+49+y^{2}-12 y+36=x^{2}+6 x+9+y^{2}-8 y+16$

$-14 x-6 x-12 y-8 y+49+36-9-16=0$

$-20 x+20 y+60=0$

$x-y-3=0$

$x-y=3$

On x-axis y is 0, so substituting y=0 we get x= 3

Hence the coordinates of point is $(3,0)$ i.e. option $(b)$ is correct