Question

Find a point on x-axis which is equidistant from $A(7,6)andB(-3,4)$

• A. $(2,0)$
• B. $(4,0)$
• C. $(3,0)$
• D. $(5,0)$

Answer 1

The correct option is C $(3,0)$

Let the point on the x-axis be $(x,0)$
Distance between $(x,0)$ and $(7,6)=\sqrt{{(7-x)}^2+{(6-0)}^2}=\sqrt{7^2+x^2-14x+36}=\sqrt{x^2-14x+85}$
Distance between $(x,0)$ and $(-3,4)=\sqrt{{(-3-x)}^2+{(4-0)}^2}=\sqrt{3^2+x^2+6x+16}=\sqrt{x^2+6x+25}$
As the point $(x,0)$ is equidistant from the two points, both the distances
calculated are equal.
$\sqrt{x^2-14x+85}=\sqrt{x^2+6x+25}$
$=>x^2-14x+85=x^2+6x+25$
$85-25=6x+14x$
$60=20x$
$x=3$
Thus, the point is $(3,0)$