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

Revision as of 17:15, 11 July 2011 by Hersheys (talk | contribs) (→‎Solution)
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Problem

AoPSWiki:Problem of the Day/July 11, 2011

Solution

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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