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 == | ||
| + | <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]] | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2022|ab=B|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Revision as of 18:39, 17 November 2022
Problem
The point 
is rotated 
counterclockwise about the point 
. What are the coordinates of its new position?
Solution 1
is 4 units west and 3 units south of 
. After a counterclockwise rotation of , which is a clockwise rotation of 
, it will end up 3 units west and 4 units north of , which is at 
.
See also
| 2022 AMC 12B (Problems • Answer Key • Resources) | |
| 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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