1971 AHSME Problems/Problem 3

Revision as of 12:21, 21 May 2016 by Hyi (talk | contribs) (Solution)

Problem 3

If the point $(x,-4)$ lies on the straight line joining the points $(0,8)$ and $(-4,0)$ in the $xy$-plane, then $x$ is equal to

$\textbf{(A) }-2\qquad \textbf{(B) }2\qquad \textbf{(C) }-8\qquad \textbf{(D) }6\qquad  \textbf{(E) }-6$

Solution

Since $(x,-4)$ is on the line $(0,8)$ and $(-4,0)$, the slope between the latter two is the same as the slope between the former two, so $\frac{-4-8}{x-0}=\frac{8-0}{0-(-4)}$. Solving for $x$, we get $x=-6$, so the answer is $\textbf{(E)}$.