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2013 AMC 12B Problems/Problem 8: Difference between revisions

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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==
==Solution 1==


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 altitutde 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>.
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>.
 
==Solution 2 - Shoelace Theorem==
 
We know lines <math>l_1</math> and <math>l_2</math> intersect at <math>B</math>, so we can solve for that point:
<cmath>3x-2y=1</cmath>
Because <math>y = 1</math> we have:
<cmath>3x-2(1) = 1</cmath>
<cmath>3x-2=1</cmath>
<cmath>3x=3</cmath>
<cmath>x = 1</cmath>
 
Thus we have <math>B = (1,1)</math>.
 
We know that the area of the triangle is <math>3</math>, so by Shoelace Theorem we have:
 
<cmath>A = \dfrac{1}{2} |(-2x+y-1) - (-2+x-y)|</cmath>
<cmath>A = \dfrac{1}{2} |-2x+y-1+2-x+y|</cmath>
<cmath>3 = \dfrac{1}{2} |-2x+y-1+2-x+y|</cmath>
<cmath>6 = |-3x+2y+1|.</cmath>
 
Thus we have two options:
 
<cmath>6 = -3x+2y+1</cmath>
<cmath>5 = -3x+2y</cmath>
 
or
 
<cmath>6 = 3x-2y-1</cmath>
<cmath>7 = 3x-2y.</cmath>
 
Now we must just find a point that satisfies <math>m_{l_3}</math> is positive.
 
Since <math>{l_3}</math> intersects <math>{l_2}</math> at <math>y=1</math>, from the second equation:
 
<cmath>7 = 3x-2(1)</cmath>
<cmath>7+2 = 3x</cmath>
<cmath>x=3</cmath>
 
so a valid point here is <math>(3,1)</math>. When calculated, the slope of <math>l_3</math> in this situation yields <math>\boxed{\textbf{(B) }\frac{3}{4}}</math>.
 
==Video Solution==
https://youtu.be/a-3CAo4CoWc
 
~Punxsutawney Phil or sugar_rush


== 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}}
[[Category:Introductory Geometry Problems]]
{{MAA Notice}}

Latest revision as of 10:52, 7 June 2025

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 1

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

Solution 2 - Shoelace Theorem

We know lines $l_1$ and $l_2$ intersect at $B$, so we can solve for that point: \[3x-2y=1\] Because $y = 1$ we have: \[3x-2(1) = 1\] \[3x-2=1\] \[3x=3\] \[x = 1\]

Thus we have $B = (1,1)$.

We know that the area of the triangle is $3$, so by Shoelace Theorem we have:

\[A = \dfrac{1}{2} |(-2x+y-1) - (-2+x-y)|\] \[A = \dfrac{1}{2} |-2x+y-1+2-x+y|\] \[3 = \dfrac{1}{2} |-2x+y-1+2-x+y|\] \[6 = |-3x+2y+1|.\]

Thus we have two options:

\[6 = -3x+2y+1\] \[5 = -3x+2y\]

or

\[6 = 3x-2y-1\] \[7 = 3x-2y.\]

Now we must just find a point that satisfies $m_{l_3}$ is positive.

Since ${l_3}$ intersects ${l_2}$ at $y=1$, from the second equation:

\[7 = 3x-2(1)\] \[7+2 = 3x\] \[x=3\]

so a valid point here is $(3,1)$. When calculated, the slope of $l_3$ in this situation yields $\boxed{\textbf{(B) }\frac{3}{4}}$.

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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