# 1963 AHSME Problems/Problem 7

## Problem

Given the four equations: $\textbf{(1)}\ 3y-2x=12 \qquad\textbf{(2)}\ -2x-3y=10 \qquad\textbf{(3)}\ 3y+2x=12 \qquad\textbf{(4)}\ 2y+3x=10$

The pair representing the perpendicular lines is: $\textbf{(A)}\ \text{(1) and (4)}\qquad \textbf{(B)}\ \text{(1) and (3)}\qquad \textbf{(C)}\ \text{(1) and (2)}\qquad \textbf{(D)}\ \text{(2) and (4)}\qquad \textbf{(E)}\ \text{(2) and (3)}$

## Solution

Write each equation in slope-intercept from since the slopes are easier to compare.

Equation $(1)$ in slope-intercept form is $y = \frac{2}{3}x+4$.

Equation $(2)$ in slope-intercept form is $y = -\frac{2}{3}x - \frac{10}{3}$.

Equation $(3)$ in slope-intercept form is $y = -\frac{2}{3}x + 4$.

Equation $(4)$ in slope-intercept form is $y = -\frac{3}{2}x + 5$.

Remember that if the two lines are perpendicular, then the product of two slopes equals $-1$. Equations $(1)$ and $(4)$ satisfy the condition, so the answer is $\boxed{\textbf{(A)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 