Difference between revisions of "2006 AMC 10A Problems/Problem 11"

(Problem)
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Which of the following describes the graph of the equation <math>\displaystyle(x+y)^2=x^2+y^2</math>?
 
Which of the following describes the graph of the equation <math>\displaystyle(x+y)^2=x^2+y^2</math>?
  
<math> \mathrm{(A) \ } the empty set\qquad \mathrm{(B) \ } one point\qquad \mathrm{(C) \ } two lines\qquad \mathrm{(D) \ } a circle\qquad \mathrm{(E) \ } the entire plane </math>
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<math> \mathrm{(A) \ } \textrm{the\,empty\,set}\qquad \mathrm{(B) \ } \textrm{one\,point}\qquad \mathrm{(C) \ } \textrm{two\,lines} \qquad \mathrm{(D) \ } \textrm{a\,circle} \qquad \mathrm{(E) \ } \textrm{the\,entire\,plane} </math>
  
 
== Solution ==
 
== Solution ==
 
Expanding the left side, we have
 
Expanding the left side, we have
  
<math>x^2+2xy+y^2=x^2+y^2\Longrightarrow 2xy=0\Longrightarrow xy=0</math>
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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>
  
There are two lines described in this graph, horizontal and vertical where x=0 and y=0. Thus, our answer is '''(C)'''
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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>.
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]

Revision as of 11:23, 31 July 2006

Problem

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

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

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 $\mathrm{(C) \ }$.

See Also