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Difference between revisions of "1961 AHSME Problems/Problem 3"
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==Problem==
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== Problem ==
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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 ''a'' is:
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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 <math>a</math> is:
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<math>\textbf{(A)}\ \pm \frac{2}{3} \qquad
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\textbf{(B)}\ -\frac{2}{3}\qquad
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\textbf{(C)}\ -\frac{3}{2} \qquad
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\textbf{(D)}\ 6\qquad
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\textbf{(E)}\ -6 </math>
  
 
==Solution==
 
==Solution==
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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 a=-6.
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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>.
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==See Also==
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{{AHSME 40p box|year=1961|num-b=2|num-a=4}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 18:33, 17 May 2018

Problem

If the graphs of $2y+x+3=0$ and $3y+ax+2=0$ are to meet at right angles, the value of $a$ is:

$\textbf{(A)}\ \pm \frac{2}{3} \qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ -\frac{3}{2} \qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ -6$

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 $a=-6$. Answer is $\boxed{\textbf{(E)}}$.

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


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