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> | ||
| − | == Solution 1 (Cartesian | + | == Solution 1 (Cartesian Plane) == |
<math>(-1,-2)</math> is 4 units west and 3 units south of <math>(3,1)</math>. Performing a counterclockwise rotation of <math>270^{\circ}</math>, which is equivalent to a clockwise rotation of <math>90^{\circ}</math>, the answer is 3 units west and 4 units north of <math>(3,1)</math>, or <math>\boxed{\textbf{(B)}\ (0,5)}</math>. | <math>(-1,-2)</math> is 4 units west and 3 units south of <math>(3,1)</math>. Performing a counterclockwise rotation of <math>270^{\circ}</math>, which is equivalent to a clockwise rotation of <math>90^{\circ}</math>, the answer is 3 units west and 4 units north of <math>(3,1)</math>, or <math>\boxed{\textbf{(B)}\ (0,5)}</math>. | ||
Revision as of 20:33, 20 November 2022
Contents
Problem
The point 
is rotated 
counterclockwise about the point 
. What are the coordinates of its new position?
Solution 1 (Cartesian Plane)
is 4 units west and 3 units south of 
. Performing a counterclockwise rotation of , which is equivalent to a clockwise rotation of 
, the answer is 3 units west and 4 units north of , or 
.
Solution 2 (Complex numbers)
We write as 
We'd like to rotate about 
which is 
in the complex plane, by an angle of 
counterclockwise.
The formula for rotating the complex number 
about the complex number 
by an angle of 
counterclockwise is given as 
Plugging in our values 
, we evaluate the expression as 
which corresponds to on the Cartesian plane.
Video Solution 1
~Education, the Study of Everything
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
