1992 AHSME Problems/Problem 12

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 $-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{-4}$ .

1992 AHSME (Problems • Answer Key • Resources)
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
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1992 AHSME Problems/Problem 12

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