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

Latest revision as of 11:21, 6 November 2021

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.

Doing some guess-and-check yields, from the second equation:

$7 = 3x-2y$ $7 = 3(3)-2(1)$ $7 = 7$

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

@@ Line 4: / Line 4: @@
 <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.
+Doing some guess-and-check yields, from the second equation:
+<cmath>7 = 3x-2y</cmath>
+<cmath>7 = 3(3)-2(1)</cmath>
+<cmath>7 = 7</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 ==

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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