Difference between revisions of "2017 AMC 10B Problems/Problem 10"

Revision as of 03:16, 4 December 2020

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 $c=\boxed{\textbf{(E)}\ 13}$

Video Solution

https://youtu.be/V4t05w7-Zd4

~savannahsolver

Video Solution 2

https://youtu.be/XRfOULUmWbY?t=582

~IceMatrix

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.

@@ Line 13: / Line 13: @@
 ~savannahsolver
+==Video Solution 2==
+https://youtu.be/XRfOULUmWbY?t=582
+~IceMatrix
 ==See Also==
 {{AMC10 box|year=2017|ab=B|num-b=9|num-a=11}}
 {{MAA Notice}}

Page

Toolbox

Search