Here, m = tan $\frac{2\pi}{3}$ = - $\sqrt{3}$ The equation of the line passing through point (0, 2) is y − 2 = − $\sqrt{3}$ (x - 0) ⇒ $\sqrt{3}x+y-2$ = 0 The slope of line parallel to $\sqrt{3}x+y-2$ = 0 is - $\sqrt{3}$ Since, it passes through (0, –2). So, the equation of line is 2 y + 2 = − $\sqrt{3}$ (x - 0) ⇒ $\sqrt{3}x+y+2$ = 0.