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...") |
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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>(E)</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>\textbf{(E)}</math>. |
Revision as of 12:21, 21 May 2016
Problem 3
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 .