Difference between revisions of "2013 AMC 10B Problems/Problem 14"
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== Alternate Solution == | == Alternate Solution == | ||
− | Following from the previous solution, <math> x^2y-xy^2 = y^2x-yx^2</math>. Then, <math>2x^2y-2xy^2=0</math>. Factoring, <math>2xy(x-y)=0</math>. Now, the solutions are obviously <math>x=0</math>, <math>y=0</math>, or <math>x=y</math>, which each correspond to a line. Thus, the answer is $\boxed{\textbf{(E)}\text{ three lines}} | + | Following from the previous solution, <math> x^2y-xy^2 = y^2x-yx^2</math>. Then, <math>2x^2y-2xy^2=0</math>. Factoring, <math>2xy(x-y)=0</math>. Now, the solutions are obviously <math>x=0</math>, <math>y=0</math>, or <math>x=y</math>, which each correspond to a line. Thus, the answer is $\boxed{\textbf{(E)}\text{ three lines}}. |
== See also == | == See also == | ||
{{AMC10 box|year=2013|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2013|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:57, 7 May 2017
Contents
[hide]Problem
Define . Which of the following describes the set of points for which ?
Solution
and . Therefore, we have the equation Factoring out a gives Factoring both sides further, . It follows that if , , or , both sides of the equation equal 0. By this, there are 3 lines (, , or ) so the answer is .
Alternate Solution
Following from the previous solution, . Then, . Factoring, . Now, the solutions are obviously , , or , which each correspond to a line. Thus, the answer is $\boxed{\textbf{(E)}\text{ three lines}}.
See also
2013 AMC 10B (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 |
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