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 $x= \frac{6}{a}$ . Similarly setting $x=0$ we find the $y-$ intercept to be $y= \frac{6}{b}$ . Then $\frac{18}{ab}=\frac{1}{2}\cdot\frac{6}{a}\cdot\frac{6}{b}$ so that $\frac{18}{ab} = 6$ , simplifying we would get $ab=3$ . Hence the answer is $\fbox{A}$ .

- $\text{[math]}$ by Kevinliu08

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1989 AHSME Problems/Problem 6

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