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Difference between revisions of "2017 AMC 10B Problems/Problem 10"
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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>
 
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>
 +
 +
==Video Solution==
 +
https://youtu.be/V4t05w7-Zd4
 +
 +
~savannahsolver
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/XRfOULUmWbY?t=582
 +
 +
~IceMatrix
  
 
==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}}

Latest revision as of 17:50, 17 January 2021

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 by TheBeautyofMath

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

~IceMatrix

See Also

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. AMC logo.png

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