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Difference between revisions of "2016 AMC 12A Problems/Problem 7"
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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>
 
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>
 +
==Diagram:==
 +
AB= <math>y=x</math> 
 +
CD= <math>y=-x</math> 
 +
EF= <math>x+y+1=0</math> 
 +
<asy>
 +
size(7cm);
 +
pair F= (5,0), E=(-1,6), D=(0,0), C=(6,0), B=(6,6), A=(0,6);
 +
draw(A--C);
 +
draw(B--D);
 +
draw(E--F);
 +
 +
label("$A$", A, dir(135));
 +
label("$B$", C, dir(-45));
 +
label("$C$", B, dir(45));
 +
label("$D$", D, dir(-135));
 +
label("$E$", E, dir(135));
 +
label("$F$", F, dir(-45));
 +
</asy>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=6|num-a=8}}
 
{{AMC12 box|year=2016|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:03, 4 February 2016

Problem

Which of these describes the graph of $x^2(x+y+1)=y^2(x+y+1)$ ?

$\textbf{(A)}\ \text{two parallel lines}\\ \qquad\textbf{(B)}\ \text{two intersecting lines}\\ \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\ \qquad\textbf{(D)}\ \text{three lines that do not all pass through a comment point}\\ \qquad\textbf{(E)}\ \text{a line and a parabola}$

Solution

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 $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$

Diagram:

AB= $y=x$ CD= $y=-x$ EF= $x+y+1=0$ [asy] size(7cm); pair F= (5,0), E=(-1,6), D=(0,0), C=(6,0), B=(6,6), A=(0,6); draw(A--C); draw(B--D); draw(E--F); label("$A$", A, dir(135)); label("$B$", C, dir(-45)); label("$C$", B, dir(45)); label("$D$", D, dir(-135)); label("$E$", E, dir(135)); label("$F$", F, dir(-45)); [/asy]

See Also

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 12 Problems and Solutions

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