2017 AMC 10B Problems/Problem 10

Problem

The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, 5)$ . What is $c$ ?

$\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$

Solution

Writing each equation in slope-intercept form, we get $y=\frac{a}{2}x-\frac{1}{2}c$ and $y=-\frac{2}{b}x-\frac{c}{b}$ . We observe the slope of each equation is $\frac{a}{2}$ and $-\frac{2}{b}$ , respectively. Because the slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$ , we see that $\frac{a}{2}=-\frac{1}{-\frac{2}{b}}$ because it is given that the two lines are perpendicular. This equation simplifies to $a=b$ .

Because $(1, 5)$ is a solution of both equations, we deduce $a \times 1-2 \times 5=c$ and $2 \times 1+b \times 5=-c$ . Because we know that $a=b$ , the equations reduce to $a-10=c$ and $2+5a=-c$ . Solving this system of equations, we get $c=\boxed{\textbf{(E)}\ 13}$

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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
All AMC 10 Problems and Solutions

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

Page

Toolbox

Search

2017 AMC 10B Problems/Problem 10

Problem

Solution

See Also