Difference between revisions of "1961 AHSME Problems/Problem 3"
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==Solution== | ==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. | 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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Revision as of 11:36, 5 July 2013
Problem
If the graphs of 
and 
are to meet at right angles, the value of a 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 a=-6.
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