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

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Writing each equation in slope-intercept form, we get <math>y=\frac{a}{2}x-\frac{1}{2}c</math> and <math>y=-\frac{2}{b}x-\frac{c}{b}</math>. We observe the slope of each equation is <math>\frac{a}{2}</math> and <math>-\frac{2}{b}</math>, respectively. Because the slope of a line perpendicular to a line with slope <math>m</math> is <math>-\frac{1}{m}</math>, we see that <math>\frac{a}{2}=-\frac{1}{-\frac{2}{b}}</math> because it is given that the two lines are perpendicular. This equation simplifies to <math>a=b</math>.
 
Writing each equation in slope-intercept form, we get <math>y=\frac{a}{2}x-\frac{1}{2}c</math> and <math>y=-\frac{2}{b}x-\frac{c}{b}</math>. We observe the slope of each equation is <math>\frac{a}{2}</math> and <math>-\frac{2}{b}</math>, respectively. Because the slope of a line perpendicular to a line with slope <math>m</math> is <math>-\frac{1}{m}</math>, we see that <math>\frac{a}{2}=-\frac{1}{-\frac{2}{b}}</math> because it is given that the two lines are perpendicular. This equation simplifies to <math>a=b</math>.
  
<math>\boxed{\textbf{(E)}\ 13}</math>
+
Because <math>(1, 5)</math> is a solution of both equations, we deduce <math>a \times 1-2 \times 5=c</math> and <math>2 \times 1+b \times 5=-c</math>. Because we know that <math>a=b</math>, the equations reduce to <math>a-10=c</math> and <math>2+5a=-c</math>. Solving this system of equations, we get <math>c=\boxed{\textbf{(E)}\ 13}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2017|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:42, 16 February 2017

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

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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

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