imc

Since the triangle is equilateral and one of the sides is a vertical line, the triangle must have a horizontal line of symmetry, and therefore the other two sides will have opposite slopes. The slope of the other given line is $\frac{\sqrt{3}}{3}$ (which is must be, given 60 degree angle of the triangle, relative to vertical) so the third must be $-\frac{\sqrt{3}}{3}$. Since this third line passes through the origin, its equation is simply $y=-\frac{\sqrt{3}}{3}x$. To find two vertices of the triangle, plug in $x=1$ to both the other equations. $y=-\frac{\sqrt{3}}{3}$ $y=1+\frac{\sqrt{3}}{3}$ We now have the coordinates of two vertices, $\left(1,-\frac{\sqrt{3}}{3}\right)$ and $\left(1,1+\frac{\sqrt{3}}{3}\right)$. The length of one side is the distance between the y-coordinates, or $1+\frac{2\sqrt{3}}{3}$. The perimeter of the triangle is thus $3\left(1+\frac{2\sqrt{3}}{3}\right)$, so the answer is $\boxed{3+2\sqrt{3}}$