Difference between revisions of "2013 AMC 12B Problems/Problem 8"
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We know that the area of the triangle is <math>3</math>, so by Shoelace Theorem we have: | 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>A = \dfrac{1}{2} |-2x+y-1+2-x+y|</cmath> | ||
<cmath>3 = \dfrac{1}{2} |-2x+y-1+2-x+y|</cmath> | <cmath>3 = \dfrac{1}{2} |-2x+y-1+2-x+y|</cmath> | ||
Line 37: | Line 37: | ||
<cmath>7 = 3x-2y.</cmath> | <cmath>7 = 3x-2y.</cmath> | ||
− | Now we must just find a point that satisfies < | + | 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: | Doing some guess-and-check yields, from the second equation: | ||
Line 45: | Line 45: | ||
<cmath>7 = 7</cmath> | <cmath>7 = 7</cmath> | ||
− | so a valid point here is < | + | 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{\dfrac{3}{4}}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 12:20, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.