Difference between revisions of "1971 AHSME Problems/Problem 3"
m (see also box, formatting tweaks) |
m (typo fix) |
||
| Line 10: | Line 10: | ||
==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>. | + | 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 == | == See Also == | ||
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 | |
| 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. 
