Difference between revisions of "2019 AMC 12B Problems/Problem 3"

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==Solution==
 
==Solution==
We can simply graph or use coordinate rules to realize that both <math>A</math> and <math>B</math> are rotated <math>180^{\circ}</math> about the origin, therefore <math>\overline{AB}</math> is rotated <math>180^{\circ}</math>, so <math>\boxed{(\text{E})}</math>
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We can simply graph or use coordinate rules to realize that both <math>A</math> and <math>B</math> are rotated <math>180^{\circ}</math> about the origin, therefore <math>\overline{AB}</math> is rotated <math>180^{\circ}</math>, so <math>\boxed{(\text{E})}</math> -Dodgers66
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:38, 14 February 2019

Problem

Which of the following rigid transformations (isometries) maps the line segment $\overline{AB}$ onto the line segment $\overline{A'B'}$ so that the image of $A(-2, 1)$ is $A'(2, -1)$ and the image of $B(-1, 4)$ is $B'(1, -4)$?

$\textbf{(A) }$ reflection in the $y$-axis

$\textbf{(B) }$ counterclockwise rotation around the origin by $90^{\circ}$

$\textbf{(C) }$ translation by 3 units to the right and 5 units down

$\textbf{(D) }$ reflection in the $x$-axis

$\textbf{(E) }$ clockwise rotation about the origin by $180^{\circ}$

Solution

We can simply graph or use coordinate rules to realize that both $A$ and $B$ are rotated $180^{\circ}$ about the origin, therefore $\overline{AB}$ is rotated $180^{\circ}$, so $\boxed{(\text{E})}$ -Dodgers66

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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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