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Difference between revisions of "2016 AMC 12A Problems/Problem 7"
 
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==Problem==
 
==Problem==
 
Which of these describes the graph of <math>x^2(x+y+1)=y^2(x+y+1)</math> ?
 
Which of these describes the graph of <math>x^2(x+y+1)=y^2(x+y+1)</math> ?
−
<math>\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}</math>  
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==Solution==
+
<math>\textbf{(A)}\ \text{two parallel lines}\\ \textbf{(B)}\ \text{two intersecting lines}\\ \textbf{(C)}\ \text{three lines that all pass through a common point}\\ \textbf{(D)}\ \text{three lines that do not all pass through a common point}\\ \textbf{(E)}\ \text{a line and a parabola}</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>
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 +
==Solution 1==
 +
 
 +
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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 +
==Solution 2==
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 +
If <math>x+y+1\neq0</math>, then dividing both sides of the equation by <math>x+y+1</math> gives us <math>x^2=y^2</math>. Rearranging and factoring, we get <math>x^2-y^2=(x+y)(x-y)=0</math>. If <math>x+y+1=0</math>, then the equation is satisfied. Thus either <math>x+y=0</math>, <math>x-y=0</math>, or <math>x+y+1=0</math>. These equations can be rearranged into the lines <math>y=-x</math>, <math>y=x</math>, and <math>y=-x-1</math>, respectively. Since these three lines are distinct, the answer is <math>\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}</math>.
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==Solution 3==
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Subtract <math>y^2(x+y+1)</math> on both sides of the equation to get <math>x^2(x+y+1)-y^2(x+y+1)=0</math>. Factoring <math>x+y+1</math> gives us <math>(x+y+1)(x^2-y^2)=(x+y+1)(x+y)(x-y)=0</math>, so either <math>x+y+1=0</math>, <math>x+y=0</math>, or <math>x-y=0</math>. Continue on with the second half of solution 2.
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==Diagram:==
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<math>AB: y=x</math>
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<math>CD: y=-x</math>
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<math>EF: x+y+1=0</math>
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<asy>
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size(7cm);
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pair F= (5,0), E=(-1,6), D=(0,0), C=(6,0), B=(6,6), A=(0,6);
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draw(A--C);
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draw(B--D);
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draw(E--F);
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label("$A$", A, dir(135));
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label("$B$", C, dir(-45));
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label("$C$", B, dir(45));
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label("$D$", D, dir(-135));
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label("$E$", E, dir(135));
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label("$F$", F, dir(-45));
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</asy>
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==See Also==
 +
{{AMC12 box|year=2016|ab=A|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 17:32, 26 December 2020

Problem

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

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

Solution 1

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

Solution 2

If $x+y+1\neq0$, then dividing both sides of the equation by $x+y+1$ gives us $x^2=y^2$. Rearranging and factoring, we get $x^2-y^2=(x+y)(x-y)=0$. If $x+y+1=0$, then the equation is satisfied. Thus either $x+y=0$, $x-y=0$, or $x+y+1=0$. These equations can be rearranged into the lines $y=-x$, $y=x$, and $y=-x-1$, respectively. Since these three lines are distinct, the answer is $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$.

Solution 3

Subtract $y^2(x+y+1)$ on both sides of the equation to get $x^2(x+y+1)-y^2(x+y+1)=0$. Factoring $x+y+1$ gives us $(x+y+1)(x^2-y^2)=(x+y+1)(x+y)(x-y)=0$, so either $x+y+1=0$, $x+y=0$, or $x-y=0$. Continue on with the second half of solution 2.

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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