2015 AMC 10A Problem 17

Since one of the sides of the equilateral triangle is a vertical line, the line of symmetry perpendicular to this side must be horizontal. This means that the slope of the third side must be opposite the slope of the second side, which would be $-\dfrac{\sqrt{3}}{3}.$ To find the perimeter, we only need to find the length of one of the sides of the triangle. We can plug in $x = 1$ into the two other equations to get the two vertices on the vertical line. The two $y$-values are $1 + \dfrac{\sqrt{3}}{3}$ and $-\dfrac{\sqrt{3}}{3}.$ Their difference is $1 + \dfrac{2\sqrt{3}}{3},$ which makes the perimeter $$3 \cdot \left(1 + \dfrac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}.$$ Thus, D is the correct answer.