The polygon(s) formed by $y=3x+2, y=-3x+2$, and $y=-2$, is (are):
$\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\ \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral}$
The points of intersection of two of the lines are $(0,2)$ and $(\pm \frac{4}{3} , -2)$, so use the Distance Formula to find the sidelengths.
Two of the side lengths are $\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}$ while one of the side lengths is $4$.  That makes the triangle isosceles, so the answer is $\boxed{\textbf{(B)}}$.