Difference between revisions of "2016 AMC 12A Problems/Problem 7"
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− | + | ==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}\\ \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> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ==Diagram:== | ||
+ | |||
+ | <math>AB: y=x</math> | ||
+ | |||
+ | <math>CD: y=-x</math> | ||
+ | |||
+ | <math>EF: 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== | ||
+ | {{AMC12 box|year=2016|ab=A|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:32, 26 December 2020
Problem
Which of these describes the graph of ?
Solution 1
The equation tells us
or
.
generates two lines
and
.
is another straight line. The only intersection of
and
is
, which is not on
. Therefore, the graph is three lines that do not have a common intersection, or
Solution 2
If , then dividing both sides of the equation by
gives us
. Rearranging and factoring, we get
. If
, then the equation is satisfied. Thus either
,
, or
. These equations can be rearranged into the lines
,
, and
, respectively. Since these three lines are distinct, the answer is
.
Solution 3
Subtract on both sides of the equation to get
. Factoring
gives us
, so either
,
, or
. Continue on with the second half of solution 2.
Diagram:
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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