1971 AHSME Problems/Problem 3

Problem

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 $\boxed{\textbf{(E) }-6}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 31 32 33 34 35
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