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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>
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== Solution 2 (Not recommended) ==
 +
It is possible to derive the formula that <math>c=-\frac{b(m^2+n^2)}{m+n}</math> for <math>ax-by = c</math> is perpendicular to <math>bx+dy=-c</math> at point <math>(m,n)</math>. Plug in <math>b=2</math>, <math>m=1</math>, and <math>n=-5</math> to get <math>c=-\frac{2 \times 26}{-4} = 13</math>, which is answer choice <math>(\text{E}) 13</math>.
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 19:23, 3 July 2025

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

Solution 2 (Not recommended)

It is possible to derive the formula that $c=-\frac{b(m^2+n^2)}{m+n}$ for $ax-by = c$ is perpendicular to $bx+dy=-c$ at point $(m,n)$. Plug in $b=2$, $m=1$, and $n=-5$ to get $c=-\frac{2 \times 26}{-4} = 13$, which is answer choice $(\text{E}) 13$.

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

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