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Difference between revisions of "2013 AMC 10B Problems/Problem 14"

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Define <math> a\clubsuit b=a^2b-ab^2 </math>. Which of the following describes the set of points <math> (x, y) </math> for which <math> x\clubsuit y=y\clubsuit x </math>?
 
Define <math> a\clubsuit b=a^2b-ab^2 </math>. Which of the following describes the set of points <math> (x, y) </math> for which <math> x\clubsuit y=y\clubsuit x </math>?
  
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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>
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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>
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== Solution==
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<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 -1 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>.

Revision as of 17:31, 21 February 2013

Problem

Define $a\clubsuit b=a^2b-ab^2$. Which of the following describes the set of points $(x, y)$ for which $x\clubsuit y=y\clubsuit x$?

$\textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel lines}\\ \qquad\textbf{(D)}\ \text{two intersecting lines}\\ \qquad\textbf{(E)}\ \text{three lines}$

Solution

$x\clubsuit y = x^2y-xy^2$ and $y\clubsuit x = y^2x-yx^2$. Therefore, we have the equation $x^2y-xy^2 = y^2x-yx^2$ Factoring out a -1 gives $x^2y-xy^2 = -(x^2y-xy^2)$ Factoring both sides further, $xy(x-y) = -xy(x-y)$. It follows that if $x=0$, $y=0$, or $(x-y)=0$, both sides of the equation equal 0. By this, there are 3 lines ($x=0$, $y=0$, or $x=y$) so the answer is $\boxed{\textbf{(E)}\text{ three lines}}$.

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