Art of Problem Solving

2006 AMC 10A Problems/Problem 11: Difference between revisions

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<math>x^2+2xy+y^2=x^2+y^2\Longrightarrow 2xy=0\Longrightarrow xy=0\Longrightarrow x = 0 \textrm{ or } y = 0</math>
<math>x^2+2xy+y^2=x^2+y^2\Longrightarrow 2xy=0\Longrightarrow xy=0\Longrightarrow x = 0 \textrm{ or } y = 0</math>


Thus there are two [[line]]s described in this graph, the horizontal line <math>y = 0</math> and the vertical line <math>x=0</math>. Thus, our answer is <math>\mathrm{(C) \ }</math>.
Thus there are two [[line]]s described in this graph, the horizontal line <math>y = 0</math> and the vertical line <math>x=0</math>. Thus, our answer is <math>\boxed{\textbf{(C) }two lines}</math>.


== See also ==
== See also ==

Revision as of 07:05, 17 December 2021

Problem

Which of the following describes the graph of the equation $(x+y)^2=x^2+y^2$?

$\textbf{(A) } \text{the\,empty\,set}\qquad \textbf{(B) } \textrm{one\,point}\qquad \textbf{(C) } \textrm{two\,lines} \qquad \textbf{(D) } \textrm{a\,circle} \qquad \textbf{(E) } \textrm{the\,entire\,plane}$

Solution

Expanding the left side, we have

$x^2+2xy+y^2=x^2+y^2\Longrightarrow 2xy=0\Longrightarrow xy=0\Longrightarrow x = 0 \textrm{ or } y = 0$

Thus there are two lines described in this graph, the horizontal line $y = 0$ and the vertical line $x=0$. Thus, our answer is $\boxed{\textbf{(C) }two lines}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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