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Difference between revisions of "1971 AHSME Problems/Problem 3"

(Created page with "== Problem 3 == If the point <math>(x,-4)</math> lies on the straight line joining the points <math>(0,8)</math> and <math>(-4,0)</math> in the <math>xy</math>-plane, then <m...")
(No difference)

Revision as of 12:19, 21 May 2016

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 $(E)$.

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