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Difference between revisions of "2016 AMC 12A Problems/Problem 7"

(Created page with " The equation <math>x^2(x+y+1)=y^2(x+y+1)</math> tells us <math>x^2=y^2</math> or <math>x+y+1=0</math>.<math>x^2=y^2</math> generates two lines <math>y=x</math> and <math>y=-x...")
 
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  The equation <math>x^2(x+y+1)=y^2(x+y+1)</math> tells us <math>x^2=y^2</math> or <math>x+y+1=0</math>.<math>x^2=y^2</math> generates two lines <math>y=x</math> and <math>y=-x</math>.<math>x+y+1=0</math> is another straight line.The only intersection of <math>y=x</math> and <math>y=-x</math> is <math>(0,0)</math>,which is not on <math>x+y+1=0</math>.Therefore,the graph is three lines that do not have a common intersection,or <math>\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}</math>
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  The equation <math>x^2(x+y+1)=y^2(x+y+1)</math> tells us <math>x^2=y^2</math> or <math>x+y+1=0</math>.<math>x^2=y^2</math> generates two lines <math>y=x</math> and <math>y=-x</math>.<math>x+y+1=0</math> is another straight line.The only intersection of <math>y=x</math> and <math>y=-x</math> is <math>(0,0)</math>,which is not on <math>x+y+1=0</math>.Therefore,the graph is three lines that do not have a common intersection,or <math>\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}</math>

Revision as of 22:56, 3 February 2016

The equation $x^2(x+y+1)=y^2(x+y+1)$ tells us $x^2=y^2$ or $x+y+1=0$.$x^2=y^2$ generates two lines $y=x$ and $y=-x$.$x+y+1=0$ is another straight line.The only intersection of $y=x$ and $y=-x$ is $(0,0)$,which is not on $x+y+1=0$.Therefore,the graph is three lines that do not have a common intersection,or $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$
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