Question

If $C$ is the midpoint of the line segment joining $A(4,0)$ and $B(0,6)$ and if O is the origin, then show that $C$ is equidistant from all the vertices of $△OAB$.

Answer 1

The midpoint of $AB$ is $C(\frac{4+0}{2},\frac{0+6}{2})=C(2,3)$
We know that the distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ is $\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
Distance between $O(0,0)$ and $C(2,3)$ is
$OC=\sqrt{{(2-0)}^2+{(3-0)}^2}=\sqrt{13}$
Distance between $A(4,0)$ and $C(2,3)$
$AC=\sqrt{{(2-4)}^2+{(3-0)}^2}=\sqrt{4+9}=\sqrt{13}$
Distance between $B(0,6)$ and $C(2,3)$
$BC=\sqrt{{(2-0)}^2+{(3-6)}^2}=\sqrt{4+9}=\sqrt{13}$
It can be observed that $OC=AC=BC$
$\therefore$ The point $C$ is equidistant from all the vertices of the $△OAB$.