Since base of an equilateral triangle lies along y-axis. So,B(0, a) and C(0, –a). Let the third vertex be A(x, 0) (Since ΔABC is an equilateral triangle and its base lies on y-axis) |BC| = |AB| = |AC| = 2a ⇒ $\sqrt{(x-0)^2+(0-a)^2}$ = 2a ⇒ $x^2+a^2$ = $4a^2$ ⇒ $x^2$ = $3a^2$ ⇒ x = ± $\sqrt{3}a$ ∴ A = $(\sqrt{3}a,0)$ or $(-\sqrt{3}a,0)$ Hence vertices of triangle are (0, a), (0, –a), and $(-\sqrt{3}a,0)$ or (0, a), (0, –a), and $(\sqrt{3}a,0)$