Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
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
⇒ (2a)2 = (OA)2 + a 2
⇒ 4a 2 – a 2 = (OA)2
⇒ (OA)2 = 3a 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.