Question

Find a point on the x-axis which is equidistant from the points $(7,6)$ and $(-3,4)$

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

Answer 1

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

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 $(-3,4)=\sqrt{{(-3-x)}^2+{(4-0)}^2}=\sqrt{9+x^2+6x+16}=\sqrt{x^2+6x+25}$
As the point $(x,0)$ is equidistant from the two points, both the distances calculated are equal.
$\sqrt{x^2-14x+85}=\sqrt{x^2+6x+25}$
$\Rightarrow x^2-14x+85=x^2+6x+25$
$\Rightarrow 85-25=6x+14x$
$\Rightarrow 60=20x$

$\Rightarrow x=3$
Thus, the point is $(3,0)$.