1986 AHSME Problems/Problem 2

Problem

If the line $L$ in the $xy$ -plane has half the slope and twice the $y$ -intercept of the line $y = \frac{2}{3} x + 4$ , then an equation for $L$ is:

$\textbf{(A)}\ y = \frac{1}{3} x + 8 \qquad \textbf{(B)}\ y = \frac{4}{3} x + 2 \qquad \textbf{(C)}\ y =\frac{1}{3}x+4\qquad\\ \textbf{(D)}\ y =\frac{4}{3}x+4\qquad \textbf{(E)}\ y =\frac{1}{3}x+2$

Solution

The original slope and $y$ -intercept are $\frac{2}{3}$ and $4$ , so the new ones are $\frac{1}{3}$ and $8$ respectively. Thus, using the slope-intercept form ( $y = mx+c$ ), the new equation is $y=\frac{1}{3}x + 8$ , which is $\boxed{A}$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1986 AHSME Problems/Problem 2

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