Difference between revisions of "2013 AMC 10B Problems/Problem 14"
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<math> \textbf{(A)}\ \text{a finite set of points}\ \qquad\textbf{(B)}\ \text{one line}\ \qquad\textbf{(C)}\ \text{two parallel lines}\ \qquad\textbf{(D)}\ \text{two intersecting lines}\ \qquad\textbf{(E)}\ \text{three lines} </math> | <math> \textbf{(A)}\ \text{a finite set of points}\ \qquad\textbf{(B)}\ \text{one line}\ \qquad\textbf{(C)}\ \text{two parallel lines}\ \qquad\textbf{(D)}\ \text{two intersecting lines}\ \qquad\textbf{(E)}\ \text{three lines} </math> | ||
− | == Solution== | + | == Solution 1== |
<math>x\clubsuit y = x^2y-xy^2 </math> and <math>y\clubsuit x = y^2x-yx^2</math>. Therefore, we have the equation <math> x^2y-xy^2 = y^2x-yx^2</math> Factoring out a <math>-1</math> gives <math> x^2y-xy^2 = -(x^2y-xy^2)</math> Factoring both sides further, <math>xy(x-y) = -xy(x-y)</math>. It follows that if <math>x=0</math>, <math>y=0</math>, or <math>(x-y)=0</math>, both sides of the equation equal 0. By this, there are 3 lines (<math>x=0</math>, <math>y=0</math>, or <math>x=y</math>) so the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. | <math>x\clubsuit y = x^2y-xy^2 </math> and <math>y\clubsuit x = y^2x-yx^2</math>. Therefore, we have the equation <math> x^2y-xy^2 = y^2x-yx^2</math> Factoring out a <math>-1</math> gives <math> x^2y-xy^2 = -(x^2y-xy^2)</math> Factoring both sides further, <math>xy(x-y) = -xy(x-y)</math>. It follows that if <math>x=0</math>, <math>y=0</math>, or <math>(x-y)=0</math>, both sides of the equation equal 0. By this, there are 3 lines (<math>x=0</math>, <math>y=0</math>, or <math>x=y</math>) so the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. | ||
− | == | + | == Solution 2== |
− | 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 <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. | + | 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 <math>\boxed{\textbf{(E)}\text{ three lines}}</math>. |
== 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}} |
Latest revision as of 14:09, 1 May 2021
Contents
[hide]Problem
Define . Which of the following describes the set of points for which ?
Solution 1
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 .
Solution 2
Following from the previous solution, . Then, . Factoring, . Now, the solutions are obviously , , or , which each correspond to a line. Thus, the answer is .
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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