2016 AMC 12A Problems/Problem 7

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$", F, dir(-45)); [/asy]$

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

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