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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, both positive (counterclockwise) and negative (clockwise).

Part 1: Rotating points by 90, 180, and 90

Let's study an example problem

We want to find the image A of the point A(3,4) under a rotation by 90 about the origin.
Let's start by visualizing the problem. Positive rotations are counterclockwise, so our rotation will look something like this:
A blank coordinate plane with a line segment where its endpoints are at the origin and a point at three, four labeled A. The point is rotated counter clockwise ninety degrees so that A prime is now in the second quadrant.
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.
A coordinate plane with a rectangle with vertices at the origin, zero, four, three, zero, and three, four which is labeled A. The x- and y- axes scale by one.
A rotation by 90 is like tipping the rectangle on its side:
A coordinate plane with a pre image rectangle with vertices at the origin, zero, four, three, zero, and three, four which is labeled A. The x- and y- axes scale by one. The rectangle is rotated ninety degrees to form the image of a rectangle with vertices at the origin, zero, three, negative four, zero, and negative four, three which is labeled A prime.
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:
Pointx-coordinatey-coordinate
A34
A43
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(x,y)=(y,x)
It turns out that this is true for any point, not just our A. Here are a few more examples:
A coordinate plane with three pre image points at eight, negative one, negative three, four, and negative three, negative six. They are rotated counter clockwise to form the image points at one, eight, negative four, negative three, and six, negative three respectively. The x- and y- axes scale by one.
Furthermore, it turns out that rotations by 180 or 90 follow similar patterns:
R(0,0),180(x,y)=(x,y)
R(0,0),90(x,y)=(y,x)
We can use these to rotate any point we want by plugging its coordinates in the appropriate equation.

Your turn!

Problem 1

Draw the image of B(7,3) under the rotation R(0,0),90.

Problem 2

Draw the image of C(5,6) under the rotation R(0,0),180.

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

Let's study an example problem

We want to find the image D of the point D(5,4) under a rotation by 270 about the origin.

Solution

Since rotating by 270 is the same as rotating by 90 three times, we can solve this graphically by performing three consecutive 90 rotations:
A coordinate plane with a pre image rectangle with vertices at the origin, zero, four, negative five, zero, and negative five, four which is labeled D. The x- and y- axes scale by one. The rectangle is rotated ninety degrees to form the image of a rectangle with vertices at the origin, zero, negative five, negative four, zero, and negative four, negative five. The rectangle is rotated ninety degrees again to form the image of a rectangle with vertices at the origin, zero, negative four, five, zero, and five, negative four. The rectangle is rotated a third time ninety degrees to form the image of a rectangle with vertices at the origin, zero, five, four, zero, and four, five which is labeled D prime.
But wait! We could just rotate by 90 instead of 270. These rotations are equivalent. Check it out:
A coordinate plane with a pre image rectangle with vertices at the origin, zero, four, negative five, zero, and negative five, four which is labeled D. The x- and y- axes scale by one. The rectangle is rotated ninety degrees clockwise to form the image of a rectangle with vertices at the origin, zero, five, four, zero, and four, five which is labeled D prime.
For the same reason, we can also use the pattern R(0,0),90(x,y)=(y,x):
R(0,0),270(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 about the origin.

Solution

A rotation by 810 is the same as two consecutive rotations by 360 followed by a rotation by 90 (because 810=2360+90).
A rotation by 360 maps every point onto itself. In other words, it doesn't change anything.
So a rotation by 810 is the same as a rotation by 90. Therefore, we can simply use the pattern R(0,0),90(x,y)=(y,x):
R(0,0),810(9,7)=(7,9)

Your turn!

Problem 1

Draw the image of E(8,6) under the rotation R(0,0),270.

Problem 2

Which rotation is equivalent to the rotation R(0,0),990?
Choose 1 answer:

Part 3: Rotating polygons

Let's study an example problem

Consider quadrilateral DEFG drawn below. Let's draw its image, DEFG, under the rotation R(0,0),270.
A cooordinate plane with a quadrilateral with vertices D at five, five, E at seven, six, F at eight, negative two, and G at two, negative two. The x- and y- axes scale by one.

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.
A cooordinate plane with a pre image quadrilateral with vertices D at five, five, E at seven, six, F at eight, negative two, and G at two, negative two. The x- and y- axes scale by one. It is rotated two hundred seventy degrees counter clockwise to form the image of the quadrilateral with vertices D prime at five, negative five, E prime at six, negative seven, F prime at negative two, negative eight, and G prime at negative two, negative two.

Your turn!

Draw the image of HIJ below, under the rotation R(0,0),90.

Want to join the conversation?

  • aqualine ultimate style avatar for user Garcia, Oswaldo
    im confused i dont get this
    (66 votes)
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    • primosaur ultimate style avatar for user Martin Schultz
      In case the algebraic method can help you:

      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)
      (55 votes)
  • boggle green style avatar for user esme.tingstrom.26
    i am actually at the end of my rope this is my downfall
    (46 votes)
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  • female robot grace style avatar for user Rey #FilmmakerForLife #EstelioVeleth.
    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
    (15 votes)
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  • duskpin sapling style avatar for user JJ
    I made a summary of the formulas if anyone dont want to use the visualize method.
    R=rotate
    R(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.
    (21 votes)
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  • starky ultimate style avatar for user wumagson
    For the first problem, why does the solution say a rotation of 90 degrees when its asking for -270
    (3 votes)
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    • piceratops ultimate style avatar for user Ian
      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!
      (32 votes)
  • blobby green style avatar for user sophia.yang
    -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
    (17 votes)
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  • blobby green style avatar for user kaurr.gg
    how do you solve for -180
    (4 votes)
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  • piceratops ultimate style avatar for user Jorge Chavolla
    Is it possible to use two different methods at once to solve an equation?
    (4 votes)
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  • duskpin tree style avatar for user Safah1
    This is very confusing because of the whole counter clock wise and clock wise, its almost as if its backwards, is there any easy way to this?
    (10 votes)
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  • piceratops sapling style avatar for user Bell Bottom
    i think i understand and then i don't understand....
    (10 votes)
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