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AoPS Wiki talk:Problem of the Day/July 11, 2011

Revision as of 17:14, 11 July 2011 by Hersheys (talk | contribs)

Problem

AoPSWiki:Problem of the Day/July 11, 2011

Solution

This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it.

For the two graphs to be perpendicular, the slopes must be the negative reciprocals of one another. That is, the product of their slopes must be $-1.$ Rewriting the first line as $y=-\frac{1}{2}x-\frac{3}{2},$ we see that its slope is $\frac{-1}{2}.$ Thus, the slope of the second line must be $2.$ Rewriting the equation of the second line, we have $y=-\frac{a}{3}x-\frac{2}{3},$ so its slope is $-\frac{a}{3}.$ Thus, $-\frac{a}{3}=2 \implies a=-6.$

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