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

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== Problem 3 ==
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== Problem ==
  
 
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 <math>x</math> is equal to
 
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 <math>x</math> is equal to
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==Solution==
 
==Solution==
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Since <math>(x,-4)</math> is on the line <math>(0,8)</math> and <math>(-4,0)</math>, the slope between the latter two is the same as the slope between the former two, so <math>\frac{-4-8}{x-0}=\frac{8-0}{0-(-4)}</math>. Solving for <math>x</math>, we get <math>x=-6</math>, so the answer is <math>\boxed{\textbf{(E) }-6}</math>.
  
Since <math>(x,-4)</math> is on the line <math>(0,8)</math> and <math>(-4,0)</math>, the slope between the latter two is the same as the slope between the former two, so <math>\frac{-4-8}{x-0}=\frac{8-0}{0-(-4)}</math>. Solving for <math>x</math>, we get <math>x=-6</math>, so the answer is <math>(E)</math>.
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=2|num-a=4}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 08:33, 1 August 2024

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
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All AHSME Problems and Solutions

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