Difference between revisions of "1971 AHSME Problems/Problem 3"
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− | == Problem | + | == 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== | ||
+ | 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>. | ||
− | + | == See Also == | |
+ | {{AHSME 35p box|year=1971|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 08:33, 1 August 2024
Problem
If the point lies on the straight line joining the points and in the -plane, then is equal to
Solution
Since is on the line and , the slope between the latter two is the same as the slope between the former two, so . Solving for , we get , so the answer is .
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.