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 (Cartesian Plane) ==
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<math>(-1,-2)</math> is <math>4</math> units west and <math>3</math> 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 <math>3</math> units west and <math>4</math> units north of <math>(3,1)</math>, or <math>\boxed{\textbf{(B)}\ (0,5)}</math>.
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~[[User: Bxiao31415 | Bxiao31415]]
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== Solution 2 (Complex Numbers) ==
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We write <math>(-1, -2)</math> as <math>-1-2i.</math> We'd like to rotate about <math>(3, 1),</math> which is <math>3+i</math> in the complex plane, by an angle of <math>270^{\circ} = \frac{3\pi}{2} \text{ rad}</math> counterclockwise.
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The formula for rotating the complex number <math>z</math> about the complex number <math>w</math> by an angle of <math>\theta</math> counterclockwise is given as <math>(z-w)e^{\theta i} + w.</math> Plugging in our values <math>z = -1-2i, w = 3+i, \theta = \frac{3\pi}{2}</math>, we evaluate the expression as <math>((-1-2i) - (3+i))e^{\frac{3\pi i}{2}} + (3+i) = (-4-3i)(-i) + (3+i) = 4i-3i^2+3+i = 4i-3+3+i = 5i,</math> which corresponds to <math>\boxed{\textbf{(B)}\ (0,5)}</math> on the Cartesian plane.
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~sirswagger21
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==Video Solution 1==
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https://youtu.be/L09yN1Y5CBI
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~Education, the Study of Everything
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==Video Solution(1-16)==
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https://youtu.be/SCwQ9jUfr0g
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~~Hayabusa1
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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}}

Latest revision as of 13:32, 5 April 2023

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 (Cartesian Plane)

$(-1,-2)$ is $4$ units west and $3$ units south of $(3,1)$. Performing a counterclockwise rotation of $270^{\circ}$, which is equivalent to a clockwise rotation of $90^{\circ}$, the answer is $3$ units west and $4$ units north of $(3,1)$, or $\boxed{\textbf{(B)}\ (0,5)}$.

~ Bxiao31415

Solution 2 (Complex Numbers)

We write $(-1, -2)$ as $-1-2i.$ We'd like to rotate about $(3, 1),$ which is $3+i$ in the complex plane, by an angle of $270^{\circ} = \frac{3\pi}{2} \text{ rad}$ counterclockwise.

The formula for rotating the complex number $z$ about the complex number $w$ by an angle of $\theta$ counterclockwise is given as $(z-w)e^{\theta i} + w.$ Plugging in our values $z = -1-2i, w = 3+i, \theta = \frac{3\pi}{2}$, we evaluate the expression as $((-1-2i) - (3+i))e^{\frac{3\pi i}{2}} + (3+i) = (-4-3i)(-i) + (3+i) = 4i-3i^2+3+i = 4i-3+3+i = 5i,$ which corresponds to $\boxed{\textbf{(B)}\ (0,5)}$ on the Cartesian plane.

~sirswagger21

Video Solution 1

https://youtu.be/L09yN1Y5CBI

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

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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