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==
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<math>[[File:Insert formula here]]</math>==Solution==
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 +
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).
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2013|ab=B|num-b=7|num-a=9}}

Revision as of 20:14, 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 $l_3$?

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

$[[File:Insert formula here]]$==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 (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