Difference between revisions of "1961 AHSME Problems/Problem 3"
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| + | ==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 ''a'' is: | 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: | ||
| + | |||
| + | ==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. | ||
Revision as of 14:01, 11 December 2012
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.
