# Difference between revisions of "1960 AHSME Problems/Problem 13"

## Problem

The polygon(s) formed by $y=3x+2, y=-3x+2$, and $y=-2$, is (are): $\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\ \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral}$

## Solution

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 $\boxed{\textbf{(B)}}$.

## See Also

 1960 AHSME (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions
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