Difference between revisions of "2014 AMC 10A Problems/Problem 14"
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Note that if the <math>y</math>-intercepts have a sum of <math>0</math>, the distance from the origin to each of the intercepts must be the same. Call this distance <math>a</math>. Since the <math>\angle PAQ = 90^\circ</math>, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is <math>\sqrt{6^2+8^2} = 10</math>, this means <math>a=10</math>, and the length of the hypotenuse is <math>2a = 20</math>. Since the <math>x</math>-coordinate of <math>A</math> is the same as the altitude to the hypotenuse, <math>[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}</math>. | Note that if the <math>y</math>-intercepts have a sum of <math>0</math>, the distance from the origin to each of the intercepts must be the same. Call this distance <math>a</math>. Since the <math>\angle PAQ = 90^\circ</math>, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is <math>\sqrt{6^2+8^2} = 10</math>, this means <math>a=10</math>, and the length of the hypotenuse is <math>2a = 20</math>. Since the <math>x</math>-coordinate of <math>A</math> is the same as the altitude to the hypotenuse, <math>[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}</math>. | ||
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+ | ==Solution 2== | ||
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+ | We can let the two lines be <cmath>y=mx+b</cmath> <cmath>y=-\frac{1}{m}x-b</cmath> This is because the lines are perpendicular, hence the <math>m</math> and <math>-\frac{1}{m}</math>, and the sum of the y-intercepts is equal to 0, hence the <math>b, -b</math>. | ||
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+ | Since both lines contain the point <math>(6,8)</math>, we can plug this into the two equations to obtain <cmath>8=6m+b</cmath> and <cmath>8=-6\frac{1}{m}-b</cmath> | ||
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+ | Adding the two equations gives <cmath>16=6m+\frac{-6}{m}</cmath> Multiplying by <math>m</math> gives <cmath>16m=6m^2-6</cmath> <cmath>\implies 6m^2-16m-6=0</cmath> <cmath>\implies 3m^2-8m-3=0</cmath> Factoring gives <cmath>(6m+2)(m-3)=0</cmath> | ||
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+ | We can just let <math>m=3</math>, since the two values of <math>m</math> do not affect our solution - one is the slope of one line and the other is the slope of the other line. | ||
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+ | Plugging <math>m=3</math> into one of our original equations, we obtain <cmath>8=6(3)+b</cmath> <cmath>\implies b=8-6(3)=-10</cmath> | ||
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+ | Since <math>\bigtriangleup APQ</math> has hypotenuse <math>2|b|=20</math> and the altutude to the hypotenuse is equal to the the x-coordinate of point <math>A</math>, or 6, the area of <math>\bigtriangleup APQ</math> is equal to <cmath>\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\ 60}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 11:44, 9 February 2014
Contents
Problem
The -intercepts, and , of two perpendicular lines intersecting at the point have a sum of zero. What is the area of ?
Solution
Note that if the -intercepts have a sum of , the distance from the origin to each of the intercepts must be the same. Call this distance . Since the , the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is , this means , and the length of the hypotenuse is . Since the -coordinate of is the same as the altitude to the hypotenuse, .
Solution 2
We can let the two lines be This is because the lines are perpendicular, hence the and , and the sum of the y-intercepts is equal to 0, hence the .
Since both lines contain the point , we can plug this into the two equations to obtain and
Adding the two equations gives Multiplying by gives Factoring gives
We can just let , since the two values of do not affect our solution - one is the slope of one line and the other is the slope of the other line.
Plugging into one of our original equations, we obtain
Since has hypotenuse and the altutude to the hypotenuse is equal to the the x-coordinate of point , or 6, the area of is equal to
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.