Difference between revisions of "1992 AHSME Problems/Problem 12"

(Created page with "== Problem == Let <math>y=mx+b</math> be the image when the line <math>x-3y+11=0</math> is reflected across the <math>x</math>-axis. The value of <math>m+b</math> is <math>\tex...")
 
(Solution)
 
Line 11: Line 11:
 
== Solution ==
 
== Solution ==
 
<math>\fbox{C}</math>
 
<math>\fbox{C}</math>
 +
First we want to put this is slope-intercept form, so we get <math>y=\dfrac{1}{3}x+\dfrac{11}{3}</math>. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since <math>m+b</math> is the sum of the slope and the y-intercept, we get <math>-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{-4}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 14:57, 22 November 2014

Problem

Let $y=mx+b$ be the image when the line $x-3y+11=0$ is reflected across the $x$-axis. The value of $m+b$ is

$\text{(A) -6} \quad \text{(B) } -5\quad \text{(C) } -4\quad \text{(D) } -3\quad \text{(E) } -2$

Solution

$\fbox{C}$ First we want to put this is slope-intercept form, so we get $y=\dfrac{1}{3}x+\dfrac{11}{3}$. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since $m+b$ is the sum of the slope and the y-intercept, we get $-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{-4}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 26 27 28 29 30
All AHSME Problems and Solutions

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

Invalid username
Login to AoPS