Difference between revisions of "1961 AHSME Problems/Problem 3"
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| − | If the graphs of <math>2y+x+3=0</math> and <math>3y+ax+2=0</math> are to meet at right angles, the value of | + | == Problem == |
| + | |||
| + | If the graphs of <math>2y+x+3=0</math> and <math>3y+ax+2=0</math> are to meet at right angles, the value of <math>a</math> is: | ||
| + | |||
| + | <math>\textbf{(A)}\ \pm \frac{2}{3} \qquad | ||
| + | \textbf{(B)}\ -\frac{2}{3}\qquad | ||
| + | \textbf{(C)}\ -\frac{3}{2} \qquad | ||
| + | \textbf{(D)}\ 6\qquad | ||
| + | \textbf{(E)}\ -6 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | The slope of the first graph is -1/2. The slope of the second is 2, since it is perpendicular, and it is also -a/3 by rearranging. Thus <math>a=-6</math>. Answer is <math>\boxed{\textbf{(E)}}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1961|num-b=2|num-a=4}} | ||
| + | |||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 18:33, 17 May 2018
Problem
If the graphs of 
and 
are to meet at right angles, the value of 
is:
Solution
The slope of the first graph is -1/2. The slope of the second is 2, since it is perpendicular, and it is also -a/3 by rearranging. Thus 
. Answer is 
.
See Also
| 1961 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
