Let ABC be the given equilateral triangle with side 2a.Accordingly, A B = B C = C A = 2 aAssume 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 △AOC, we obtain(AC)2=(OA)2+(OC)2⇒(2a)2=(OA)2+a2⇒4a2−a2=(OA)2⇒(OA)2=3a2⇒OA=3​​ ∴ Coordinates of point A=(±3​a,0) Thus, the vertices of the given equilateral triangle are (0, a),(0,–a), and (3​a,0) or (0,a),(0,−a), and(−3​a,0)