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

(N/A) Let $ABC$ be the equilateral triangle with side $2a$.
Since the base $BC$ lies along the $y$-axis with its mid-point at the origin $O(0,0)$,the coordinates of $B$ and $C$ are $(0, a)$ and $(0, -a)$ (or vice versa).
The altitude from vertex $A$ to the base $BC$ lies along the $x$-axis because the altitude of an equilateral triangle is perpendicular to the base.
In $\Delta AOC$,where $O$ is the origin,$AC = 2a$ and $OC = a$.
Using the Pythagorean theorem: $(AC)^2 = (OA)^2 + (OC)^2$
$(2a)^2 = (OA)^2 + a^2$
$4a^2 = (OA)^2 + a^2$
$(OA)^2 = 3a^2$
$OA = \sqrt{3}a$
Since $A$ lies on the $x$-axis,its coordinates are $(\sqrt{3}a, 0)$ or $(-\sqrt{3}a, 0)$.
Thus,the vertices of the triangle are $(0, a), (0, -a), (\sqrt{3}a, 0)$ or $(0, a), (0, -a), (-\sqrt{3}a, 0)$.