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

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1963 AHSME Problems/Problem 7

Problem

Solution

See Also