Difference between revisions of "2022 AMC 12B Problems/Problem 5"

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<math>\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)</math>
 
<math>\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)</math>
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== Solution 1 ==
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<math>(-1,-2)</math> is 4 units west and 3 units south of <math>(3,1)</math>. After a counterclockwise rotation of <math>270^{\circ}</math>, which is a clockwise rotation of <math>90^{\circ}</math>, it will end up 3 units west and 4 units north of <math>(3,1)</math>, which is at <math>\fbox{(0,5) B}</math>.
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~[[User: Bxiao31415 | Bxiao31415]]
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== See also ==
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{{AMC12 box|year=2022|ab=B|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 17:39, 17 November 2022

Problem

The point $(-1, -2)$ is rotated $270^{\circ}$ counterclockwise about the point $(3, 1)$. What are the coordinates of its new position?

$\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)$

Solution 1

$(-1,-2)$ is 4 units west and 3 units south of $(3,1)$. After a counterclockwise rotation of $270^{\circ}$, which is a clockwise rotation of $90^{\circ}$, it will end up 3 units west and 4 units north of $(3,1)$, which is at $\fbox{(0,5) B}$.

~ Bxiao31415

See also

2022 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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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