1989 AHSME Problems/Problem 6

Problem

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 $\fbox{A}$.

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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