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

(Fixed grammar, improved clarity, and removed an irrelevant attribution)
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<math>\textbf{(B) } </math> counterclockwise rotation around the origin by <math>90^{\circ}</math>
 
<math>\textbf{(B) } </math> counterclockwise rotation around the origin by <math>90^{\circ}</math>
  
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<math>\textbf{(C) } </math> translation by 3 units to the right and 5 units down
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<math>\textbf{(C) } </math> translation by <math>3</math> units to the right and <math>5</math> units down
  
 
<math>\textbf{(D) } </math> reflection in the <math>x</math>-axis
 
<math>\textbf{(D) } </math> reflection in the <math>x</math>-axis

Revision as of 19:33, 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 $\boxed{(\text{E})}$.

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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