1992 AHSME Problems/Problem 12

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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
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