Art of Problem Solving
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

2016 AMC 12A Problems/Problem 7: Difference between revisions

← Older editNewer edit →
No edit summary
Arpitr20 (talk | contribs)
editor
32 edits
Line 7: Line 7:


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

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_7&oldid=75391"