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>? | ||
− | <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 | + | <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== | ||
+ | <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 . Which of the following describes the set of points for which ?
Solution
and . Therefore, we have the equation Factoring out a -1 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 .