Difference between revisions of "2013 AMC 12B Problems/Problem 8"

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==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>?
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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 18:04, 22 February 2013

Problem

Line $l_1$ has equation $3x - 2y = 1$ and goes through $A = (-1, -2)$. Line $l_2$ has equation $y = 1$ and meets line $l_1$ at point $B$. Line $l_3$ has positive slope, goes through point $A$, and meets $l_2$ at point $C$. The area of $\triangle ABC$ is $3$. What is the slope of $l_3$?

$\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{3}{4} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{4}{3} \qquad \textbf{(E)}\ \frac{3}{2}$

Solution

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
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