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

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

Revision as of 16:51, 21 December 2020

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 altitude 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 $\frac{|1-(-2)|}{|3-(-1)|}=$ $\boxed{\textbf{(B) }\frac{3}{4}}$.

Video Solution

https://youtu.be/a-3CAo4CoWc

~Punxsutawney Phil or sugar_rush

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

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