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### Course: High school geometry>Unit 1

Lesson 4: Rotations

# Rotating shapes about the origin by multiples of 90°

Learn how to draw the image of a given shape under a given rotation about the origin by any multiple of 90°.

## 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 ${A}^{\prime }$ of the point $A\left(3,4\right)$ 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:
Cool, we estimated ${A}^{\prime }$ 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 $A$.
A rotation by ${90}^{\circ }$ is like tipping the rectangle on its side:
Now we see that the image of $A\left(3,4\right)$ under the rotation is ${A}^{\prime }\left(-4,3\right)$.
Notice it's easier to rotate the points that lie on the axes, and these help us find the image of $A$:
Point$\left(3,0\right)$$\left(0,4\right)$$\left(3,4\right)$
Image$\left(0,3\right)$$\left(-4,0\right)$$\left(-4,3\right)$

### Solution method 2: The algebraic approach

Let's take a closer look at points $A$ and ${A}^{\prime }$:
Point$x$-coordinate$y$-coordinate
$A$$3$$4$
${A}^{\prime }$$-4$$3$
Notice an interesting phenomenon: The $x$-coordinate of $A$ became the $y$-coordinate of ${A}^{\prime }$, and the opposite of the $y$-coordinate of $A$ became the $x$-coordinate of ${A}^{\prime }$.
We can represent this mathematically as follows:
${R}_{\left(0,0\right),{90}^{\circ }}\left(x,y\right)=\left(-y,x\right)$
It turns out that this is true for any point, not just our $A$. Here are a few more examples:
Furthermore, it turns out that rotations by ${180}^{\circ }$ or $-{90}^{\circ }$ follow similar patterns:
${R}_{\left(0,0\right),{180}^{\circ }}\left(x,y\right)=\left(-x,-y\right)$
${R}_{\left(0,0\right),-{90}^{\circ }}\left(x,y\right)=\left(y,-x\right)$
We can use these to rotate any point we want by plugging its coordinates in the appropriate equation.

#### Problem 1

Draw the image of $B\left(-7,-3\right)$ under the rotation ${R}_{\left(0,0\right),-{90}^{\circ }}$.

#### Problem 2

Draw the image of $C\left(5,-6\right)$ under the rotation ${R}_{\left(0,0\right),{180}^{\circ }}$.

### 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 ${D}^{\prime }$ of the point $D\left(-5,4\right)$ 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:
But wait! We could just rotate by $-{90}^{\circ }$ instead of ${270}^{\circ }$. These rotations are equivalent. Check it out:
For the same reason, we can also use the pattern ${R}_{\left(0,0\right),-{90}^{\circ }}\left(x,y\right)=\left(y,-x\right)$:
${R}_{\left(0,0\right),{270}^{\circ }}\left(-5,4\right)=\left(4,5\right)$

### Let's study one more example problem

We want to find the image of $\left(-9,-7\right)$ 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 $810=2\cdot 360+90$).
A rotation by ${360}^{\circ }$ maps every point onto itself. In other words, it doesn't change anything.
So a rotation by ${810}^{\circ }$ is the same as a rotation by ${90}^{\circ }$. Therefore, we can simply use the pattern ${R}_{\left(0,0\right),{90}^{\circ }}\left(x,y\right)=\left(-y,x\right)$:
${R}_{\left(0,0\right),{810}^{\circ }}\left(-9,-7\right)=\left(7,-9\right)$

#### Problem 1

Draw the image of $E\left(8,6\right)$ under the rotation ${R}_{\left(0,0\right),-{270}^{\circ }}$.

#### Problem 2

Which rotation is equivalent to the rotation ${R}_{\left(0,0\right),-{990}^{\circ }}$?

## Part 3: Rotating polygons

### Let's study an example problem

Consider quadrilateral $DEFG$ drawn below. Let's draw its image, ${D}^{\prime }{E}^{\prime }{F}^{\prime }{G}^{\prime }$, under the rotation ${R}_{\left(0,0\right),{270}^{\circ }}$.

### 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.

Draw the image of $\mathrm{△}HIJ$ below, under the rotation ${R}_{\left(0,0\right),{90}^{\circ }}$.

## Want to join the conversation?

• im confused i dont get this

Rotating by 90 degrees:
If you have a point on (2, 1) and rotate it by 90 degrees, it will end up at (-1, 2)
When you rotate by 90 degrees, you take your original X and Y, swap them, and make Y negative.
So from 0 degrees you take (x, y), swap them, and make y negative (-y, x) and then you have made a 90 degree rotation.

What if we rotate another 90 degrees? Same thing.
Our point is at (-1, 2) so when we rotate it 90 degrees, it will be at (-2, -1)
X and Y swaps, and Y becomes negative.

What about 90 degrees again? Same thing! But remember that a negative and a negative gives a positive so when we swap X and Y, and make Y negative, Y actually becomes positive.
Our point is as (-2, -1) so when we rotate it 90 degrees, it will be at (1, -2)

Another 90 degrees will bring us back where we started.
From (1, -2) to (2, 1)

Rotating by -90 degrees:
If you understand everything so far, then rotating by -90 degrees should be no issue for you.
We do the same thing, except X becomes a negative instead of Y.
So from (x, y) to (y, -x)

Rotating by 180 degrees:
If you have a point on (2, 1) and rotate it by 180 degrees, it will end up at (-2, -1)
When you rotate by 180 degrees, you take your original x and y, and make them negative.
So from 0 degrees you take (x, y) and make them negative (-x, -y) and then you've made a 180 degree rotation.

Remember! A negative and a negative gives a positive! So if we rotate another 180 degrees we go from (-2, -1) to (2, 1)
And if we have another point like (-3, 2) and rotate it 180 degrees, it will end up on (3, -2)
• i am actually at the end of my rope this is my downfall
• real this makes me pluck out my eyebrows
• I made a summary of the formulas if anyone dont want to use the visualize method.
R=rotateR(0,0)90°/-270°[counterclock wise/clockwise](x,y)➔(-y,x)R(0,0)180°/-180°[counterclock wise/clock wise](x,y)➔(-x,-y)R(0,0)-90°/270°[clock wise/counterclock wise](x,y)➔(y,-x)R(0,0)360°/-360°(x,y)=(x,y)

In all honesty, there are only two that need to be memorized. R 90°/-270°and R -90°/270°since depends on the direction it rotates the x and y value sometimes turns into negative, but 100% of the time the x and y swaps. The rest of two you just need to know how it works, R 180°/-180 is just reflected through origin so you dont have to swap x and y, all you have to do is add - sign to them. And R360°/-360° is just rotating back to where it was, so nothing is changed.
• Anyone have any tips for visualization? I am having a really hard time seeing a triangle and where the point should go in my head. Any suggestions are appreciated very much!
Thank you,
Clarebugg
• you can try thinking of it as a mountain
the bottom is the 2 points that stretch out and the top is the peak.
there are many different ways to think about it.
you are problem-solving by trying to visualize.
• -90 and 270:
(y, -x)
reverse, neg x

90 and -270
(-y,x)
reverse, neg y

(-)180
(-x,-y)
easiest to remember - just negify everything
• For the first problem, why does the solution say a rotation of 90 degrees when its asking for -270
• A rotation of 90 degrees is the same thing as -270 degrees. They are called coterminal angles. A full way around a circle is 360 degrees, right? So you can find an angle by adding 360. -270 + 360 = 90

Hope this helps!
• i think i understand and then i don't understand....
• Amazing. Just Amazing