The base of an equilateral triangle with side $2 a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Let $ABC$ be the given equilateral triangle with side $2 a$.

Accordingly, $A B=B C=C A=2 a$

Assume that base $BC$ lies along the $y-$ axis such that the mid-point of $BC$ is at the origin.

i.e., $BO = OC = a ,$ where $O$ is the origin.

Now, it is clear that the coordinates of point $C$ are $(0, a),$ while the coordinates of point $B$ are $(0,-a).$

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex $A$ lies on the $y-$ axis.

On applying Pythagoras theorem to $\Delta$ $AOC$, we obtain

$(A C)^{2}=(O A)^{2}+(O C)^{2}$

$\Rightarrow(2 a)^{2}=(O A)^{2}+a^{2}$

$\Rightarrow 4 a^{2}-a^{2}=(O A)^{2}$

$\Rightarrow(O A)^{2}=3 a^{2}$

$\Rightarrow O A=\sqrt{3} a$

$\therefore$ Coordinates of point $A=(\pm \sqrt{3} a, 0)$

Thus, the vertices of the given equilateral triangle are $(0, a),(0,-a)$, and$(\sqrt{3} a, 0)$ or $(0, a),(0,-a)$, and $(-\sqrt{3} a, 0)$