Introduction
In this article we will practice the art of rotating shapes. Mathematically speaking, we will learn how to draw the image of a given shape under a given rotation.
This article focuses on rotations by multiples of $90^\circ$, both positive (counterclockwise) and negative (clockwise).
Part 1: Rotating points by $90^\circ$, $180^\circ$, and $-90^\circ$
Let's study an example problem
We want to find the image $A'$ of the point $A(3,4)$ under a rotation by $90^\circ$ about the origin.
Let's start by visualizing the problem. Positive rotations are counterclockwise, so our rotation will look something like this:
[[☃ image 1]]
Cool, we estimated $A'$ visually. But now we need to find exact coordinates. There are two ways to do this.
Solution method 1: The visual approach
We can imagine a rectangle that has one vertex at the origin and the opposite vertex at $A$.
[[☃ image 1]]
A rotation by $90^\circ$ is like tipping the rectangle on its side:
[[☃ image 2]]
Now we see that the image of $A(3,4)$ under the rotation is $A'(-4,3)$.
Notice it's easier to rotate the points that lie on the axes, and these help us find the image of $A$:
| Point | $(3,0)$ | $(0,4)$ | $(3,4)$ |
|---|---|---|---|
| Image | $(0,3)$ | $(-4,0)$ | $(-4,3)$ |
Solution method 2: The algebraic approach
Let's take a closer look at points $A$ and $A'$:
| Point | $x$-coordinate | $y$-coordinate |
|---|---|---|
| $\large A$ | $\large\tealD{3}$ | $\large\purpleC{4}$ |
| $\large A'$ | $\large-\purpleC{4}$ | $\large\tealD{3}$ |
Notice an interesting phenomenon: The $x$-coordinate of $A$ became the $y$-coordinate of $A'$, and the opposite of the $y$-coordinate of $A$ became the $x$-coordinate of $A'$.
We can represent this mathematically as follows:
$R_{(0,0),90^\circ}(\tealD{x},\purpleC{y})=(-\purpleC{y},\tealD{x})$
It turns out that this is true for any point, not just our $A$. Here are a few more examples:
[[☃ image 1]]
Furthermore, it turns out that rotations by $180^\circ$ or $-90^\circ$ follow similar patterns:
$R_{(0,0),180^\circ}(\tealD{x},\purpleC{y})=(-\tealD{x},-\purpleC{y})$
$R_{(0,0),-90^\circ}(\tealD{x},\purpleC{y})=(\purpleC{y},-\tealD{x})$
We can use these to rotate any point we want by plugging its coordinates in the appropriate equation.
Your turn!
Problem 1
[[☃ graded-group 1]]
[[☃ explanation 1]]
Problem 2
[[☃ graded-group 1]]
[[☃ explanation 1]]
Graphical method vs. algebraic method
In general, everyone is free to choose which of the two methods to use. Different strokes for different folks!
The algebraic method takes less work and less time, but you need to remember those patterns. The graphical method is always at your disposal, but it might take you longer to solve.
Part 2: Extending to any multiple of $90^\circ$
Let's study an example problem
We want to find the image $D'$ of the point $D(-5,4)$ under a rotation by $270^\circ$ about the origin.
Solution
Since rotating by $270^\circ$ is the same as rotating by $90^\circ$ three times, we can solve this graphically by performing three consecutive $90^\circ$ rotations:
[[☃ image 1]]
But wait! We could just rotate by $-90^\circ$ instead of $270^\circ$. These rotations are equivalent. Check it out:
[[☃ image 2]]
For the same reason, we can also use the pattern $R_{(0,0),-90^\circ}(\tealD{x},\purpleC{y})=(\purpleC{y},-\tealD{x})$:
$R_{(0,0),270^\circ}(-5,4)=(4,5)$
Let's study one more example problem
We want to find the image of $(-9,-7)$ under a rotation by $810^\circ$ about the origin.
Solution
A rotation by $810^\circ$ is the same as two consecutive rotations by $360^\circ$ followed by a rotation by $90^\circ$ (because $810=2\cdot360+90$).
A rotation by $360^\circ$ maps every point onto itself. In other words, it doesn't change anything. [[☃ explanation 1]]
So a rotation by $810^\circ$ is the same as a rotation by $90^\circ$. Therefore, we can simply use the pattern $R_{(0,0),90^\circ}(\tealD{x},\purpleC{y})=(-\purpleC{y},\tealD{x})$:
$R_{(0,0),810^\circ}(-9,-7)=(7,-9)$
Your turn!
Problem 1
[[☃ graded-group 1]]
[[☃ explanation 1]]
Problem 2
[[☃ graded-group 1]]
[[☃ explanation 1]]
Part 3: Rotating polygons
Let's study an example problem
Consider quadrilateral $DEFG$ drawn below. Let's draw its image, $D'E'F'G'$, under the rotation $R_{(0,0),270^\circ}$.
[[☃ image 1]]
Solution
Similar to translations, when we rotate a polygon, all we need is to perform the rotation on all of the vertices, and then we can connect the images of the vertices to find the image of the polygon.
[[☃ image 1]]
[[☃ explanation 1]]
Your turn!
[[☃ graded-group 1]]
[[☃ explanation 1]]