Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 12A Problems/Problem 7"
Line 1: Line 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>
+
==Problem==
 +
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>
 +
==Solution==
 +
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 22:08, 3 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 $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_7&oldid=75241"