Difference between revisions of "2013 AMC 12B Problems/Problem 8"
m (→Problem) |
|||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | Line <math>l_1</math> has equation <math>3x - 2y = 1</math> and goes through <math>A = (-1, -2)</math>. Line <math>l_2</math> has equation <math>y = 1</math> and meets line <math>l_1</math> at point <math>B</math>. Line <math>l_3</math> has positive slope, goes through point <math>A</math>, and meets <math>l_2</math> at point <math>C</math>. The area of <math>\triangle ABC</math> is 3. What is the slope of <math>l_3</math>? | + | Line <math>l_1</math> has equation <math>3x - 2y = 1</math> and goes through <math>A = (-1, -2)</math>. Line <math>l_2</math> has equation <math>y = 1</math> and meets line <math>l_1</math> at point <math>B</math>. Line <math>l_3</math> has positive slope, goes through point <math>A</math>, and meets <math>l_2</math> at point <math>C</math>. The area of <math>\triangle ABC</math> is <math>3</math>. What is the slope of <math>l_3</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> | <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> | ||
Revision as of 17:04, 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
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 | |
