The point on x-axis which is equidistant from the points (7, 6), (-3, 4) is

Solution:

Any point on X-axis is represented as (x,0)

Distance of this point from (7,6) = $$\sqrt{\ \left(7-x\right)^2\ +\left(6-0\right)^2}$$= 

Distance of this point from (-3,4) = $$\sqrt{\ \left(-3-x\right)^2\ +\left(4-0\right)^2}$$

Both the distances should be same.

$$\sqrt{\ \left(7-x\right)^2\ +\left(6-0\right)^2}$$ = $$\sqrt{\left(\left(-3-x\right)^2\ \ +\left(4\right)^2\right)\ }$$

$$\sqrt{\ \left(49+x^2\ -14x\right)+36}\ =\ \sqrt{\ 9+x^2+6x+16}$$

$$\left(49+x^2\ -14x\right)+36\ =\ 9+x^2+6x+16$$

$$60=20x$$

$$x\ =\ 3$$

So, point = $$\left(3,0\right)$$ Answer