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== | + | ==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 | + | 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 == | == See also == |
Latest revision as of 12:21, 6 November 2021
Problem
Line has equation
and goes through
. Line
has equation
and meets line
at point
. Line
has positive slope, goes through point
, and meets
at point
. The area of
is
. What is the slope of
?
Solution 1
Line has the equation
when rearranged. Substituting
for
, we find that line
will meet this line at point
, which is point
. We call
the base and the altitude from
to the line connecting
and
,
, the height. The altitude has length
, and the area of
. Since
,
. Because
has positive slope, it will meet
to the right of
, and the point
to the right of
is
.
passes through
and
, and thus has slope
.
Solution 2 - Shoelace Theorem
We know lines and
intersect at
, so we can solve for that point:
Because
we have:
Thus we have .
We know that the area of the triangle is , so by Shoelace Theorem we have:
Thus we have two options:
or
Now we must just find a point that satisfies is positive.
Doing some guess-and-check yields, from the second equation:
so a valid point here is . When calculated, the slope of
in this situation yields
.
Video Solution
~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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