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 \\(\\Delta\\)AOC, we obtain
\\(\\begin{array}{l}{(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}\\end{array}\\)
\\(\ herefore \ ext { Coordinates of point } \\mathrm{A}=\\left( \\begin{array}{c}{ \\pm \\sqrt{3} a, 0 )}\\end{array}\\right.\\)
\\(\\begin{array}{l}{\ ext { Thus, the vertices of the given equilateral triangle are }(0, a),(0,-a), \ ext { and }(\\sqrt{3} a, 0) \ ext { or }} \\\\ {(0, \ ext { a), }(0,-a), \ ext { and }}\\end{array}\\) \\((-\\sqrt{3} a, 0)\\).