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Difference between revisions of "1989 AHSME Problems/Problem 6"
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==Solution==
 
==Solution==
 
Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x=6/a</math>. Similarly setting <math>x=0</math> we find the <math>y-</math>intercept to be <math>y=6/b</math>. Then <math>6=(1/2)(6/a)(6/b)</math> so that <math>ab=3</math>. Hence the answer is <math>\mathrm{(A)}</math>.
 
Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x=6/a</math>. Similarly setting <math>x=0</math> we find the <math>y-</math>intercept to be <math>y=6/b</math>. Then <math>6=(1/2)(6/a)(6/b)</math> so that <math>ab=3</math>. Hence the answer is <math>\mathrm{(A)}</math>.
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{{MAA Notice}}

Revision as of 12:47, 5 July 2013

If $a,b>0$ and the triangle in the first quadrant bounded by the co-ordinate axes and the graph of $ax+by=6$ has area 6, then $ab=$

$\mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 }$

Solution

Setting $y=0$ we have that the $x-$intercept of the line is $x=6/a$. Similarly setting $x=0$ we find the $y-$intercept to be $y=6/b$. Then $6=(1/2)(6/a)(6/b)$ so that $ab=3$. Hence the answer is $\mathrm{(A)}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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