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

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==Solution==
 
==Solution==
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We can simply graph the points, or use coordinate geometry, to realize that both <math>A'</math> and <math>B'</math> are, respectively, obtained by rotating <math>A</math> and <math>B</math> by <math>180^{\circ}</math> about the origin. Hence the rotation by <math>180^{\circ}</math> about the origin maps the line segment <math>\overline{AB}</math> to the line segment <math>\overline{A'B'}</math>, so the answer is <math>\boxed{(\text{E})}</math>.
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We can simply graph the points, or use coordinate geometry, to realize that both <math>A'</math> and <math>B'</math> are, respectively, obtained by rotating <math>A</math> and <math>B</math> by <math>180^{\circ}</math> about the origin. Hence the rotation by <math>180^{\circ}</math> about the origin maps the line segment <math>\overline{AB}</math> to the line segment <math>\overline{A'B'}</math>, so the answer is <math>\boxed{\textbf{(E)}}</math>.
  
 
==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 18:59, 18 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 the points, or use coordinate geometry, to realize that both $A'$ and $B'$ are, respectively, obtained by rotating $A$ and $B$ by $180^{\circ}$ about the origin. Hence the rotation by $180^{\circ}$ about the origin maps the line segment $\overline{AB}$ to the line segment $\overline{A'B'}$, so the answer is $\boxed{\textbf{(E)}}$.

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
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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