Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AHSME Problems/Problem 6"

(Created page with "If <math>a,b>0</math> and the triangle in the first quadrant bounded by the co-ordinate axes and the graph of <math>ax+by=6</math> has area 6, then <math>ab=</math> <math> \math...")
 
Line 2: Line 2:
  
 
<math> \mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 }  </math>
 
<math> \mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 }  </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>A</math>.

Revision as of 21:04, 29 February 2012

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 $ab=$

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

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 $A$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_6&oldid=45240"