Difference between revisions of "2013 AMC 12B Problems/Problem 8"
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==Solution== | ==Solution== | ||
− | 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 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>. |
== See also == | == See also == |
Revision as of 12:48, 18 February 2014
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
Line has the equation
when rearranged. Substituting
for
,we find that line
will meet this line at point
, which is point B. We call
the base and the altitude from A to the line connecting B and C,
, 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
.
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.