# 2022 AMC 12B Problems/Problem 5

## 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 $\boxed{\textbf{(B)}\ (0,5)}$.

## 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

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## Video Solution(1-16)

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 2022 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 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