**Solution:**

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

Accordingly, AB = BC = CA = 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 ΔAOC, we obtain

(AC)^{2} = (OA)^{2} + (OC)^{2}

⇒ (2*a*)^{2} = (OA)^{2} + *a* ^{2}

⇒ 4*a* ^{2} – *a* ^{2} = (OA)^{2}

⇒ (OA)^{2} = 3*a* ^{2}

⇒ OA =

∴Coordinates of point A =

Thus, the vertices of the given equilateral triangle are (0, *a*), (0, –*a*), and or (0, *a*), (0, –*a*), and.