Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 11, 2011"
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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 <math>-1.</math> Rewriting the first line as <math>y=-\frac{1}{2}x-\frac{3}{2},</math> we see that its slope is <math>\frac{-1}{2}.</math> Thus, the slope of the second line must be <math>2.</math> Rewriting the equation of the second line, we have <math>y=-\frac{a}{3}x-\frac{2}{3},</math> so its slope is <math>-\frac{a}{3}.</math> Thus, <math>-\frac{a}{3}=2 \implies a=-6.</math> | 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 <math>-1.</math> Rewriting the first line as <math>y=-\frac{1}{2}x-\frac{3}{2},</math> we see that its slope is <math>\frac{-1}{2}.</math> Thus, the slope of the second line must be <math>2.</math> Rewriting the equation of the second line, we have <math>y=-\frac{a}{3}x-\frac{2}{3},</math> so its slope is <math>-\frac{a}{3}.</math> Thus, <math>-\frac{a}{3}=2 \implies a=-6.</math> | ||
Latest revision as of 18:15, 11 July 2011
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 
Rewriting the first line as 
we see that its slope is 
Thus, the slope of the second line must be 
Rewriting the equation of the second line, we have 
so its slope is 
Thus, 
