Difference between revisions of "2013 AMC 12B Problems/Problem 8"
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<math>\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{3}{4} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{4}{3} \qquad \textbf{(E)}\ \frac{3}{2}</math> | <math>\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{3}{4} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{4}{3} \qquad \textbf{(E)}\ \frac{3}{2}</math> | ||
| − | + | ==Solution== | |
Line ''l''1 has the equation y=3/2x-1/2 when rearranged. Line''l''2 will meet this line at point B(1,1). | Line ''l''1 has the equation y=3/2x-1/2 when rearranged. Line''l''2 will meet this line at point B(1,1). | ||
Revision as of 19:16, 22 February 2013
Problem
Line 
has equation 
and goes through 
. Line 
has equation 
and meets line at point 
. Line 
has positive slope, goes through point 
, and meets at point 
. The area of 
is 
. What is the slope of ?
Solution
Line l1 has the equation y=3/2x-1/2 when rearranged. Linel2 will meet this line at point B(1,1).
See also
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
