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

Revision as of 08:58, 7 April 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

Line $l_1$ has the equation $y=3x/2-1/2$ when rearranged. Substituting $1$ for $y$ ,we find that line $l_2$ will meet this line at point $(1,1)$ , which is point B. We call $\overline{BC}$ the base and the altitutde from A to the line connecting B and C, $y=-1$ , the height. The altitude has length $|-2-1|=3$ , and the area of $\triangle{ABC}=3$ . Since $A={bh}/2$ , $b=2$ . Because $l_3$ has positive slope, it will meet $l_2$ to the right of $B$ , and the point $2$ to the right of $B$ is $(3,1)$ . $l_3$ passes through $(-1,-2)$ and $(3,1)$ , and thus has slope $\frac{|1-(-2)|}{|3-(-1)|}=$ $\boxed{\textbf{(B) }\frac{3}{4}}$ .

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

@@ Line 10: / Line 10: @@
 == See also ==
 {{AMC12 box|year=2013|ab=B|num-b=7|num-a=9}}
+[[Category:Introductory Geometry Problems]]

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

Revision as of 08:58, 7 April 2013

Problem

Solution

See also