{{ 'ml-label-loading-course' | message }}

{{ tocSubheader }}

{{ 'ml-toc-proceed-mlc' | message }}

{{ 'ml-toc-proceed-tbs' | message }}

An error ocurred, try again later!

Chapter {{ article.chapter.number }}

{{ article.number }}. # {{ article.displayTitle }}

{{ article.intro.summary }}

Show less Show more Lesson Settings & Tools

| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount }} |

| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount }} |

| {{ 'ml-lesson-time-estimation' | message }} |

When learning about transformations, rotations were mentioned and used briefly. In this lesson, rotations will be studied more deeply by analyzing the relationships between a point $Q$ and its image $Q_{′}$ under a rotation.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Explore

In the following applet, points $A,$ $B,$ $C,$ and $D$ can be moved and rotated about point $P.$

Make an arrangement of points and rotate them about $P.$ Then, find the distance between $P$ and each preimage, and the distance between $P$ and each image.

$PAPC =?=? PA_{′}PC_{′} =?=? PBPD =?=? PB_{′}PD_{′} =?=? $

Next, make a different arrangement, rotate the points, and find the distances. What can be said about rotations? If needed, make more arrangements and rotations.Explore

Perform a rotation to $△JKL$ about point $P.$ Use a protractor to find the measure of $∠JPJ_{′},$ $∠KPK_{′},$ and $∠LPL_{′}.$

Change the triangle and perform the same process. What can be said about rotations? Does the conclusion depend on the position of $P$ relative to the triangle?

Discussion

The first exploration shows the preimage and the image of a point under a rotation are the same distance from the center of rotation. That is, the image moves along a circle passing through the preimage and centered at the center of rotation.

The second exploration shows that after performing a rotation, the angles formed by each preimage, the center of rotation, and the corresponding image all have the same measure. With these properties in mind, rotations can be properly defined.

Discussion

Identifying whether one figure is the image of another figure under rotation can be difficult. A key aspect to observe is whether the center of rotation is the same distance from an image as it is from its preimage.

Concept

A rotation is a transformation in which a figure is turned about a fixed point $P.$ The number of degrees the figure rotates $α_{∘}$ is the angle of rotation. The fixed point $P$ is called the center of rotation. Rotations map every point $A$ in the plane to its image $A_{′}$ such that one of the following statements is satisfied.

- If $A$ is the center of rotation, then $A$ and $A_{′}$ are the same point.
- If $A$ is not the center of rotation, then $A$ and $A_{′}$ are equidistant from $P$, with $∠APA_{′}$ measuring $α_{∘}.$

Since rotations preserve side lengths and angle measures, they are rigid motions.

Example

Consider the following three triangles and a point $P.$ Use the given measuring tool to find the distance from each vertex to $P$ and the angles formed by each preimage, the point $P,$ and the corresponding image.
If one of the triangles is a rotation of $△ABC$ about $P,$ what is the measure of the angle of rotation? ### Hint

### Solution

Which of the triangles, $△A_{′}B_{′}C_{′}$ or $△A_{′′}B_{′′}C_{′′}$, is the image of $△ABC$ under a rotation about $P?$

{"type":"choice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.946332em;vertical-align:-0.19444em;\"><\/span><span class=\"mord\">\u25b3<\/span><span class=\"mord\"><span class=\"mord mathdefault\">A<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.751892em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.05017em;\">B<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.751892em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">C<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.751892em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.946332em;vertical-align:-0.19444em;\"><\/span><span class=\"mord\">\u25b3<\/span><span class=\"mord\"><span class=\"mord mathdefault\">A<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.751892em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2032<\/span><span class=\"mord mtight\">\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.05017em;\">B<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.751892em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2032<\/span><span class=\"mord mtight\">\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">C<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.751892em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2032<\/span><span class=\"mord mtight\">\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","None","Both"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["150","210"]}}

Remember, after performing a rotation, the preimage and the image of a point are the same distance from the center of rotation. The angle of rotation is formed by a preimage, the center of rotation, and the corresponding image.

Remember that, after performing a rotation, the preimage and the image of a point are the same distance from the center of rotation. Then, start by finding the distances between each vertex and $P.$

Notice that the vertices of $△A_{′}B_{′}C_{′}$ are further from $P$ than the vertices of $△ABC.$ Consequently, $△A_{′}B_{′}C_{′}$ cannot be the image of $△ABC$ under a rotation about $P.$ Next, find and compare the measures of $∠APA_{′},$ $∠BPB_{′},$ and $∠CPC_{′}.$

As can be seen, $∠APA_{′},$ $∠BPB_{′},$ and $∠CPC_{′}$ have all the same measure, which is $150_{∘}$ when measured counterclockwise or $210_{∘}$ when measured clockwise. Therefore, $△A_{′′}B_{′′}C_{′′}$ is the image of $△ABC$ under a rotation about $P.$ The angle of rotation is either $150_{∘}$ or $210_{∘}.$

Discussion

Rotations can be performed by hand with the help of a straightedge, a compass, and a protractor.

To rotate point $A$ about point $P$ by an angle of $130_{∘}$ measured counterclockwise, follow these five steps.

1

Draw $PA$

2

Place the Protractor

Place the center of the protractor on $P$ and align it with $PA.$

The protractor is placed as illustrated above when the rotation is counterclockwise. If the rotation has to be done clockwise, the protractor needs to be placed as follows.

3

Mark the Desired Angle

Locate the corresponding measure on the protractor and make a small mark. In this case, the mark will be made at $130_{∘}.$

4

Draw a Ray

Using the straightedge, draw a ray with starting point $P$ that passes through the mark made in the previous step.

5

Draw Point $A_{′}$

Place the compass tip on $P$ and open it to the distance between $P$ and $A.$ Without changing this setting and keeping the point of the compass at $P,$ draw a small arc centered at $P$ that intersects the ray drawn before.

The intersection of the ray and the arc is the image $A_{′}$ after the give rotation.

Notice that this method of construction has also confirmed that $PA$ is congruent to $PA_{′}.$

Example

On a geometry test, Ignacio was asked to perform a $70_{∘}$ counterclockwise rotation to $△ABC$ about point $P.$

Draw $△ABC$ and its image under this rotation.

To rotate $△ABC$, the rotation can be performed on each vertex, one at a time. For example, start by rotating $A.$ To do so, first draw $PA$ using a straightedge.

Then, place the center of the protractor on $P$ and align it with $PA$ in such a way that the rotation is counterclockwise. Then, make a small mark at $70_{∘}.$

Next, draw a ray with starting point $P$ that passes through the mark made before.

Finally, place the compass tip on $P$ and open the compass to the distance between $P$ and $A.$ With this setting, make an arc that intersects the ray. The intersection point is the image of $A$ under the rotation.

Vertices $B$ and $C$ can be rotated following the same steps.

Finally, the image of $△ABC$ under the given rotation is the triangle formed by $A_{′},$ $B_{′},$ and $C_{′}.$

Pop Quiz

Discussion

In the coordinate plane, there is a particular relationship between the coordinates of a point and those of its image after a counterclockwise rotation about the origin. This relation occurs when the angle of rotation is either $90_{∘},$ $180_{∘},$ or $270_{∘}.$ Try to figure it out by using the following applet.

From the diagram, the following relations can be set.

- The image of $(a,b)$ under a $90_{∘}$ rotation about the origin is $(-b,a).$
- The image of $(a,b)$ under a $180_{∘}$ rotation about the origin is $(-a,-b).$
- The image of $(a,b)$ under a $270_{∘}$ rotation about the origin is $(b,-a).$

Example

Given a figure and its image under a rotation, the following theorem can be used to find the center of rotation.

If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment. |

With this theorem in mind, consider the following example. In the diagram, quadrilateral $A_{′}B_{′}C_{′}D_{′}$ is the image of $ABCD$ under a certain rotation.

Find the center and angle of rotation.

**Angle of Rotation:** $120_{∘}$ clockwise or $240_{∘}$ counterclockwise.

**Graph:**

Remember that the center of rotation is equidistant from the preimage and the image of each vertex. Use the Converse Perpendicular Bisector Theorem. The center is the intersection point between two perpendicular bisectors.

The first step is to find the center of rotation. Remember, by definition, a point and its image under a rotation are the same distance from the center.

The center of rotation is equidistant from a point and its image.

Therefore, by the Converse of the Perpendicular Bisector Theorem, the center lies on the perpendicular bisector of $AA_{′},$ for instance. Then, with the aid of a compass and a straightedge, start by constructing the perpendicular bisector of this segment.

To determine the center's exact position, draw a second segment joining a vertex and its image, for example, $DD_{′}.$ Then, draw the perpendicular bisector of this segment. The intersection between both perpendicular bisectors is the center of rotation.

Notice that drawing only two perpendicular bisectors is enough to find the center of rotation because all will intersect at the same point. Since the sense of rotation was not specified, both measures will be found using a protractor.

The angle of rotation is either $240_{∘}$ counterclockwise or $120_{∘}$ clockwise.

Pop Quiz

In the following applet, the solid triangle is the image of the dashed triangle under a certain counterclockwise rotation. Place the point where the center of the rotation should be. Then, select the appropriate angle of rotation.
### Extra

About the Applet

If the point is not placed close enough to the center of rotation, when the Check Answer

button is pushed, a red area is highlighted indicating the region where the center of rotation is located.

Example

Recall that rotations are transformations and that transformations can be composed. Therefore, it is possible to have a composition of two or more rotations. On a geometry exercise, the following two rotations are given.

- $R_{1}:$ $90_{∘}$ counterclockwise rotation about $C_{1}(0,0).$
- $R_{2}:$ $180_{∘}$ counterclockwise rotation about $C_{2}(0,-1).$

LaShay has to perform both rotations to $△ABC,$ one after the other, but the book does not indicate the composition's order.

According to the book, the correct coordinates of $B_{′′}$ are $(2,-1).$ Knowing this, help LaShay to find the coordinates of $C_{′′}$ and draw $△A_{′′}B_{′′}C_{′′}.${"type":"text","form":{"type":"point2d","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"decimal":false,"function":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text1":"4","text2":"1"}}

Start by applying any of the given two rotations. For example, apply $R_{1}$ first. Notice it is a $90_{∘}$ counterclockwise rotation about the origin. Then, it maps $(a,b)$ onto $(-b,a).$ Knowing this, the image of $A,$ $B,$ and $C$ can be quickly found.

$A(-4,1)B(-1,2)C(-3,4) →→→ A_{′}(-1,-4)B_{′}(-2,-1)C_{′}(-4,-3) $

Therefore, the image of $△ABC$ under $R_{1}$ looks as follows.
Next, apply $R_{2}$ to $△A_{′}B_{′}C_{′}.$ Remember, it is a $180_{∘}$ counterclockwise rotation about $C_{2}(0,-1).$
Since the coordinates of $B_{′′}$ are the same as the book says, the rotations were applied in the correct order. Consequently, the coordinates of $C_{′′}$ are $(4,1).$ Notice that applying the rotations in the reverse order would produce a different image.

Discussion

Think about the preimage and the final image obtained by LaShay in the previous example. Is it possible to map $△ABC$ onto $△A_{′′}B_{′′}C_{′′}$ performing only one rotation? The answer is yes! A $270_{∘}$ counterclockwise rotation about $(-1,-1)$ does it.
*except* if the sum of their angles of rotation is a multiple of $360_{∘}.$ Investigate what happens in this case by using the following applet.

In general, the composition of two rotations is a rotation,

- Place the centers $A$ and $B$ at any place within the square formed by the axes and the point $(-1,1).$
- The slider on the left performs a rotation about point $A$ and the slider on the right performs a rotation about point $B.$
- Perform rotations such that the sum of their angles of rotation equals $360_{∘}.$
- Compare the preimage and the image.

As might be checked, if the angles of rotation add up to $360_{∘},$ the final image is not a rotation of the original preimage but a translation.

Closure

In real life, there are plenty of situations where rotations can be appreciated. For instance, take a look at a door.

- To open a door, the handle must be rotated.
- To lock or unlock it, a key must be inserted and rotated.
- While the door is opening, it rotates around the hinges.

Loading content