Similarity Transformations

Transformations

Isometries

Reflections Translations Rotations Glide Reflections

Answers to exercises

1 Transformations

In geometry, the process by which one can effect a change to a set of points is called a transformation. A transformation requires an initial set or initial object and some very specific instructions about what to do to it. After being applied to the initial set, a transformation produces an image set. We will focus on transformations that result in location changes, size changes, or both. Transformations are sometimes called mappings from an initial set to an image set or a rigid motion in the plane.

We will find that some notation is necessary; this will be developed as we move along. In general we’ll specify a transformation with an initial or a few letters and give some pertinent information about the transformation and then we’ll specify the initial set or object. The fact that this will look like functional notation is not an accident.

For example, a transformation, B, involving a reference point C to be used on the initial set that is the segment AB would be specified:

B(C) (AB).

This is read “do transformation B with reference to C to the initial set AB”.

The image set of AB after B(C) is accomplished will be marked with coordinates that have prime marks on them: A’B’. A’ is the image of A and B’ is the image of B. All of the points in the original set have image points, but only the ones important enough to label will be marked in this fashion.

It is possible to perform a series of transformations on an initial set and have interim image sets appear along the way to a final image set. If we want multiple transformations, we’ll list them from the last applied to the first applied, left to right separated by a small hollow dot “° “. This is called composition of transformations and the hollow dot is read “after”. For example suppose we want 3 transformations: B1, B2, and B3 and we want them performed in numerical order to initial set ABC with reference to a line L, we’ll write a composition string:

B3 ° B2 ° B1 (L)(ABC)

This is read “do B3 after B2 after B1 with reference to L to initial set ABC”.

With compositions, we may indicate reference information in the composition string or in writing before the composition string.

With transformations, there is a concern about what properties of the initial set stay the same (“preserved by the transformation”) and what changes. I will be discussed these issues as we go along.

2 Similarity Transformations

A transformation that changes the distance between points by a fixed amount is called a similarity transformation or a dilation*. The fixed amount of the change is called the scale factor.

Each similarity transformation requires an initial set, a scale factor, k, that is a real number, and a point of dilation, P. A single similarity transformation will be denoted as follows:

S(k, P)(initial set).

Often, in text books, the instructions are not presented in this format but are discussed in a sentence like: “Dilate the set MN by a scale factor of 5 about the point Q”. We will write the following instead of the sentence:

S(5, Q)(MN).

It is concise and imitates the format of the transformations that follow.

*This is what the similarity transform is called under the transformation menu in Sketchpad. You indicate the point of dilation by double-clicking on it first and then clicking on key points of the initial set.

3 Illustration with a segment and a center of dilation at the midpoint

Given the line segment AB with the distance from A to B of 3 cm. If we want to change the distance from A to B by a factor of 3 using the midpoint of the segment, we write S(3, midpoint)(AB).

The image set is a segment A’B’ with length 9cm. Note that the midpoint of the image set is still halfway in between A’ and B’ and is right on top of the initial midpoint .

AB = 3.00 cm A'B' = 9.00 cm

m idpointB' = 4.50 cm m idpointA' = 4.50 cm

m idpoint

Note that the midpoint of the original segment is halfway between A and B and the image midpoint’ is halfway between A’ and B’.

We say that similarity transformations “preserve betweeness” because if a point is located between two points in the original set, it’s image point will be located proportionally between the image points of the two bracketing image points.

4 Illustration with a segment and a center of dilation not on the segment

Suppose we have another segment RS with the length of RS = 6 cm and we want to dilate it about point C, not on the segment, by a factor of 0.5: S(0.5, C)(RS). Let’s look at at a couple of pictures to see what happens.

Here is the initial set and the point of dilation:

Here are the initial set and the image set with C after the dilation is complete. Note that the transformed end points, the image endpoints, are called R’ and S’.

R'S' = 3.00 cm

5 Note that the initial set, segment RS, and the image set, segment R’S’, are parallel line segments. Note, too, that the image segment length is ½ the length of the initial segment, precisely the original length times the scale factor.

Furthermore, the initial and image segment endpoints are on straight segments that emanate from the point of dilation to the endpoints. Line segment CR is called a “projection line” as is line segment CS. You will always have projection lines when the point of dilation is not a point of the initial set.

The location of the image set is closer to C because the scale factor is a fraction thus reducing it in size.

What if k were 1.5? Where would the image set be with respect to the initial set and C? (ans: on the other side of RS further from the point of dilation…extend the projection lines outward and sketch in R’’S’’ from S(1.5, C) (RS).

Dilation Exercise 1:

A. Using the point P and segment AB below: S(1.5, P)(AB). Show projection lines. B. Then using the midpoint of AB: S(1.5, midpoint)(AB). Note the lack of projection lines with this second dilation. This is because the point of projection is part of the initial set and not a distance from the initial set.

How do the lengths of the two images compare?

A P

midpoint

BA = 6.00 c m

6 Illustration of preserving betweeness on a segment:

Here is a segment with A – C – B, in fact AC = .75 AB. S(2, D)(AB) and see what happens to the relative location of C’.

AB = 4.00 cm AC = 3.00 cm

D B

C B' A'B' = 8.00 cm A'C' = 6.00 cm C'

A' Not only is the betweeness relationship preserved but the relative proportions between the parts of the segment are preserved: C’ is .75 A’B’.

The initial point set is the segment AB and the image point set, the post-transformation set, is A’B’. These two segments are parallel. This is characteristic of dilations or similarity transformations with a point of dilation separate from the initial set.

Connect DAA’ and DBB’ with projection lines; do you see the nested triangles with a parallel side? Can you see why this transformation is called a “similarity” transformation?

Exercise with a right triangle, showing the creation of similar triangles

C Suppose we have a 3cm – 4cm – 5cm right triangle, ABC. 5cm 3cm

A 4cm B 2 We want to use a scale factor of and as center of dilation the vertex of the right angle, 3 point A. Our instructions are

2 S( , A)(ABC). 3

Here’s a picture of what happens.

C A. What is the relationship between ABC, the original set, and AC’B’ the image set? C'

A B' B

B. What is the perimeter of the original triangle?

C. What is the perimeter of the image triangle?

8 D. What is the area of the original triangle?

E. What is the area of the image triangle?

F. What is the relationship between CB and C'B' ? How do you know?

G. What are the measures of the angles of AB’C’?

We say that similarity transformations “preserve” angle measure, because while the location and length of a side of a figure may change, the measures of the angles between sides of a figure does not change when it undergoes a similarity transformation.

H. If we put the original set in the Cartesian coordinate plane with vertex A at the origin, what are the coordinates of vertex C and vertex B?

I. What are the geometric coordinates of the image vertices C’ and B’?

2 J. Do these new coordinates reflect a similarity transformation of ? 3

9 Isosceles Right Triangles Exercise:

Suppose we have an isosceles right triangle with the congruent legs measuring 3 cm and we want to perform a similarity transformation using a point C that is nearby like this.

A B

S(2, C)(ARB) gives us

A B

A' B'

A. What is the length of side A’R’? B’R’?

10 The corresponding sides under the transformation are parallel. B. What does this tell you about the relationship between the two triangles?

Yes they are similar, so the angle measures of corresponding angles are congruent.

If you connect the image, the original and C along each vertex, you get straight lines that look like this:

A B

A' B'

These lines are called perspective lines or projection lines.

Note that C is sometimes called “the vanishing point”. Can you guess why?

11 Here’s a hint: S(0.25, C)(ARS) is going to show up with vertices A’’, B’’, and R’’… look where it is located.

C k = 0.25 R''

A" B" original set R

R' k = 2 A B

A' B'

Note that the line segments A"R" , A'R' , and AR are parallel. This is true for each of the other triples of corresponding sides as well. Since the sides of the original triangle are parallel to the corresponding sides of each image, we know that all three triangles are similar.

C. What would we use as a scale factor to go from the smallest triangle back to the original triangle? From the largest triangle back to the original triangle?

D. Are the angle measures in corresponding angles congruent for all three triangles?

E. What is the area of each triangle? How are the areas related to the scale factors?

Note that Similarity transformations preserve angle measure. An angle that has a certain measure initially, will still have that measure after a dilation. That fact is why an initial triangle and it’s image after a dilation are similar triangles.

12 Example – the old fashioned way

Here is a line segment PQ , and a center of dilation, C. We will sketch in a similarity transformation with a scale factor of 2 using old fashioned tools.

The first step is to sketch in dotted lines from C through each endpoint and further; use a straight edge or the edge of a sheet of paper. We need to extend the lines beyond the segment to the right because the scale factor is more than 1. (Do you see why? How far would we have to draw the lines if the scale factor were 1/3?)

Your sketch should look like this:

13 Now, using the edge of a different sheet of paper, mark off the distance from P to Q with tiny precise tick marks. Then, using one of the tick marks as a midpoint, mark off a length that is 2PQ. You should have 3 tiny tick marks in a row on the edge; the distance from the leftmost to the rightmost is 2PQ. Call the leftmost tick mark P’ and the rightmost Q’; the one in the center is the midpoint.

The edge of the paper should have tick marks like this – only yours should be to scale with the segment.

P’ Q’

Now gently fold the paper with the original PQ on it so that ray CP is on top of ray CQ. This puts a small fold on the midpoint of the original segment. Using a straightedge, draw in the midpoint ray, extending out like the projection lines.

Line up the midpoint of the original segment with the midpoint of the segment on the edge of the paper ( P'Q' ). Gently pull the paper away from C, keeping the paper edge parallel to the original segment and the midpoint on the midpoint ray.. When P’ and Q’ are lined up on the dotted lines, sketch in the image set on them using the piece of paper as a straight edge. This line is precisely twice the length of the original segment and is the image set of the similarity transformation S(2, C)(PQ).

I must point out that this procedure is FAR easier in Sketchpad!

14 example

Now try S(2, D)(ABC) on your own.

15 An illustration of orientation preservation.

Suppose you have an asymmetric figure: ABCDE.

Notice that when you start at A and move clockwise around the figure you have labeled points in alphabetical order.

Now: S(P, 2)(ABCDE)

C' D'

D C

B' E' B E P

Notice that the “flag” part of the figure is still to the right and that you can start with A’ and go clockwise around the figures and touch the points in the same order:

ABCDE::A’B’C’D’E’

Symmetry transformations preserve the orientation of a figure.

To get the idea of this, let’s move ahead to a reflection:

16 Here is an example of a figure that underwent a transformation that did not preserve it’s orientation. Is this a similarity transformation? How do you know – hint start at A and go clockwise around the shape…start at A’ and go clockwise around the shape. Are the image points in the same order as the initial untransformed points? No. Reflections do not “preserve orientation”’ they are “orientation reversing”.

D C' C D'

B E E' B'

A A'

The only reason to note that Similarity transformations are orientation preserving is because there are transformations that do not preserve orientation. Noting what is preserved can help you backtrack from an image to an initial set and figure out what kind of transformation has happened.

17 Dilation Exercise 2:

Using Sketchpad,

S(2, center C)(circle). How do you explain the location of the image circle center?

Then do

S(2, P)(circle).

What is different about each sketch? What is the same about each sketch?

P C C'

CC' = 2.00 cm CP = 5.00 cm

Construction tips: place C and use Translate (2/1, 0) for C’; construct a circle with center and point. Translate from C to get P.

18 Summary

A similarity transformation (a dilation)

requires a point of dilation, a scale factor, and an initial set

S(k, P)(initial set)

preserves angle measure preserves betweeness preserves orientation

changes distances from point to point from initial to image and may change the location of the initial set completely (as long as k is not 1)

19 Isometries

A transformation that preserves the distance between any two points is an isometry (from Greek: iso for same; metry for measure).

There is only one simple isometry: a reflection.

The other three commonly mentioned (translation, rotation, and glide reflection) are compositions of reflections.

Properties of an isometry:

preserves distances preserves angle measures preserves betweeness

changes location

20 Reflections

A reflection is an isometry that takes the original set and creates a mirror image of it across a specified line of reflection. Note that this may also be called a mapping or a motion.

It is necessary to use a totally asymmetric figure to understand fully the nature of a reflection. Suppose we have the following original set ABCD (on the left below) and L is our given line of reflection.

L C

The notation that to produce a mirror image of ABCD across line L is R(L) (ABCD).

D A' C L

The mirror image is A’B’C’D’

21 Note that when you start with A and read off the points going counterclockwise you get ABCD. On the image, starting with A’ and going counterclockwise you get a different order A’D’C’B’ and no permutation* of the points will reproduce the original point order. This means that the orientation of the object has been changed. This orientation reversal is a key feature of a reflection. We call the reflection “odd” to indicate the reversal of orientation. We’ll use “odd” in this manual but be aware that there are several terms in general use.

Reflections are orientation reversing motions.

One way to illustrate orientation reversal is to hold your left hand in front of a mirror and notice that what you see as the image is a right hand…the image is related to the original set but has its orientation irretrievably reversed.

* A permutation is a reordering of a list of objects in a way that does not allow items on the list to “jump”. For example, “A, B, C, D” is an ordered list; “D, A, B, C” is a permutation of the list while “A, D, B, C” is not. You must always move the last letter to the front of the list to get another permutation. This is a special use of the word. If the transformation is orientation preserving, then the list of high profile image points taken in a counterclockwise direction will be a permutation of the list of the corresponding initial points taken in a counterclockwise position. vnet video: Permutations

Permutation exercise:

Given the points 1234, list all the permutations of them.

22 Let’s look at another fact about reflections:

The line of reflection, L, is the perpendicular bisector of any line joining an original point and it’s image. This relationship is true for all points and image points under a reflection.

C' GD' = 1.33 cm

D' DG = 1.33 cm G

D E A' C L

B F m EFA' = 90.00

Initial Exercises:

A. Is this a reflection? Why or why not?

23 B. Sketch in the line of reflection How do you know you’re right?

C' D

D' B' A E

Reflection exercise:

R(A’B)(C”CAC’).

Illustrate that the transformation is orientation reversing and show the corresponding measures on the image

AA' = 3.00 cm m AA'B = 90.00 m C'CC" = 60.00 CC' = 3.00 cm B C

A A'

24 Properties of an reflection:

R(L) (initial set)

preserves distances preserves angle measures preserves betweeness

changes location

reverses orientation*

*alternate language for this behavior: odd or opposite. Any one of these three descriptives means that the orientation is changed by the motion.

25 The other three transformations that produce a change in location are compositions of reflections. We will look at them from this standpoint and from the traditional view that each can be specified as an individual transformation (motion or mapping).

Translations and Rotations

The composition of two reflections produces a translation or a rotation.

A translation is the composition* of two reflections across parallel lines of reflection. A rotation is the composition of two reflections across intersecting lines of reflection. \

Both of these types of mappings are “even” or orientation preserving.

Glide Reflections

A glide reflection is the composition* of three reflections.

It is an orientation reversing motion in the plane.

* Remember that a composition is just doing two transformations, one right after the other. It is listing with the last or final transformation on the left and the first on the right of a tiny hollow dot that is read “after”. See the introduction for a review.

26 Translations

A composition of two reflections using parallel lines produces a translation. We may specify the motion as a composition or a translation in our notation.

To specify a translation as a single transformation, we will say

T(magnitude, angle)(initial set)

The magnitude is the length of the motion along the angle and the angle is the angle of elevation: the number of degrees up from the horizontal (or down with a negative). All of the lengths in this section will be centimeters so we will simply specify a number. If you are in a situation where inches and meters – mixed units of measure – are available you should specify a unit of measure as well.

To specify the same translation as a composition of reflections, we well say

R2 (L 2 )° R 1 (L 1 )(initial set)

And you will have to show the two parallel lines on the sketch.

The image set will be the SAME if the specifications are done properly.

For example:

If you have a point set ABC, you may specify a translation of 4cm at 0 angle of elevation: T(4 cm,0°)(ABC).

4 cm is the magnitude of the translation and it is the distance from an initial point to its corresponding image point once the translation is completed.

Let’s draw in a couple of straight lines joining initial points and image points to see how the information in the parenthesis after “T” shows up.

27 B B'

C C'

A A'

Note that if you connect A and A’ or B and B’ the length of the connecting segment is 4 cm. Note further that these connecting segments are parallel.

Is this an orientation preserving or reversing transformation?

Start at A and go clockwise around the initial set. Start at the image of A and go clockwise around the image set. What do you find?

All translations are the composition of two reflections about parallel lines of reflection. Let’s decompose the above translation to see this. I have to insert two parallel lines in the right places to make this work. The placement of the lines is NOT unique; other choices will work just as well. It’s just that you can’t put them in arbitrarily. Pick the first one and then put the second one where it will make the same result.

We may specify:

R2(L2)° R1(L1)(ABC) = T(4 cm, 0) and achieve exactly the same translation.

28 L1 L2 m ABC' = 54.35 B B' m A'B'C' = 54.35 BC' = 0.85 cm

C'C C' B'C' = 0.85 cm

A A'

The distance from A to A’ is 4 cm.

We have exactly the same initial and final image sets for both transformations so they are equal. The interim image – what you get after the first reflection, before you’re done is shown in dotted lines.

You can be somewhat relaxed about where the first line of reflection goes. It just needs to be perpendicular to a line at the angle of elevation off the horizontal. The second line of reflection should be the midpoint between an interim image and where you want the final image to be, also perpendicular to a line at the angle of elevation off the horizontal.

Note that the interim image – shown with dotted lines is orientation reversed, but the final image has the same orientation as the initial set.

Translations are direct motions or “even” transformation. When the transformation is completed, the image has the same orientation as the initial set.

vnet video: Translations – how to get the reflection lines

Here is T(3 cm, 60°)(AB) where AB is a circle.

29 T(3 cm, 60°)(AB) exercise:

Illustrate how the “3cm” and the “60” affected the initial set to produce the final set.

Then show proposed lines of reflection for this transformation.

I’ve put a spare couple of circle sets below for you to try to put in the pair of parallel reflection lines on your own. Try first BEFORE you peek at the answer. Note that you might be successful with lines entirely different from the ones I chose. There are LOTS and lots of pairs of parallel reflection lines that will work.

A' A'

B' B'

A A

B B

T(6 cm,45)(ABCD) exercise:

30 Show how the 45 can be seen. Show two lines of reflection that could produce T(6 cm,45°)(ABC).

Translation exercise:

31 Using the provided lines of reflection, show the translation. Approximate the measurements that you would use to specify the motion as a translation.

B B'''

Properties of an Translation:

32 T(magnitude, angle)(initial set) preserves distances preserves angle measures preserves betweeness

changes location

preserves orientation (is a direct or even transformation)

can be written as the composition of reflections about two parallel lines

33 Rotations

Composing two reflections using intersecting lines of reflection produces a rotation.

R2(L2)° R1(L1)(initial set) = Rot (angle, point)(initial set)

This is read: do Reflection 2 about Line 2 after you do Reflection 1 about Line 1 to the initial set.

You may also say: Rotate the initial set by “angle” degrees about the given point. These behaviors will result in the same image set if your set-ups are correct.

Note that we need an angle and a point that is the center of the rotation. Reflections are about lines; rotations are about points.

I’ll use the point called “center” on the line as my point for the rotation of ABCDE. Note that ABCDE is asymmetric.

Here’s an initial set ABCDE and we’ll apply the following transformation: Rot (center, 180°) (ABCDE).

D C A'

E B center B' E'

C' D' A

What do you observe about this image set? It is an isometry – the angle measures and side lengths are unchanged. But the orientation: same or reversed?

34 Note that ABCDE is the order when you start at A and go counterclockwise around the initial set.

Let’s start at A’ and go counterclockwise around the image set. What do you get? What happens if you start at D? (You get a permutation of the transformed initial points.)

This fact is what is meant when we say that a rotation preserves orientation. Note, too, that this means that a rotation is what we call a direct or even transformation – it preserves orientation just like translations do.

So you can check the orientation of the image to distinguish whether or not there’s a reflection or a rotation. (Which is to say, do we have one reflection or a composition of two reflections about crossed lines).

How would we decompose this rotation into two reflections?

The claim is that two reflections around intersecting lines of reflection produce a rotation. Here’s the rotation from above recast as a composition of reflections:

Note that I renamed “center” as C; it’s really the same point.

interim object k

C final object

initial object

35 First reflect the initial object about line j to get the interim object. Then reflect the interim object about line k to get the final object. ( “Object” is an alternative to saying “set”; some audiences don’t really have enough of a background in set theory to see ABCDE as a point set.)

Notice that each line of reflection serves as a perpendicular bisector of lines joining the original and image points. That’s because each is line of reflection. Note, too that they intersect. This helps distinguish rotations from translations. Both rotations and translations preserve orientation, but the reflection lines for translations are parallel while the lines for rotations intersect.

So R2 (k)° R 1 ( j)(ABCDE) is the same as a 180 rotation about C. Note that C is the point of intersection of the reflection lines. The reflection lines are NOT unique…there are other lines that would have worked just as well – they, too, would intersect at C, though. The two above just happen to be perpendicular, too – in general, the lines are not perpendicular.

How can you tell whether you have a single reflection or a rotation? Sometimes if the object has the right sort of symmetry you actually can’t tell…using a circle as your object, for example, isn’t helpful. There’s a homework exercise about this point.

Let’s look at our initial set ABCDE again and see what happens if we delete C and reflect about the line instead.

Here’s a single reflection about the line:

D C C' D'

B' E B E'

A A'

Note that the reflection is an isometry – all the angle measures and segment lengths are identical in the initial set and the image set – and note that the orientation of the initial set is reversed. Do you see that the pictures really are different?

36 vnet: Rotations – how to get the reflection lines

Illustration:

Rotate the following set about point O an angle of 120.

Rot(O, 120°)(ABCD)

O C

Here’s the result: Note how to see the 120 rotation -- you can sketch in the angles!

O C

B m DOD' = 120.00 m BOB' = 120.00 m AOA' = 120.00 A B' D'

37 Now show the change as the composition of two reflections:

C O A' B

A B' D'

I chose a line of reflection THROUGH O. Then I put in the angle bisector between where an interim point A is and where I want it to end up (A’) using O as the vertex of the angle to be bisected. L2 is the angle bisector of angle A’interimOA’image and I used it as my second line of reflection.

L2 L1

C O

A B' D'

38 Distinguishing exercise:

Is the following a rotation or a reflection?

What about this one?

39 Rotation exercise:

The initial set is BCDE. This shows a rotation about point A. What is the angle of the rotation? . (use a protractor!) Show lines of reflection that would create it. Give the instructions for doing it in both composed reflections and rotation notation.

E' A C D

C' B' D' E

40 Properties of an Rotation:

Rot (angle, point)(initial set)

preserves distances preserves angle measures preserves betweeness

changes location

preserves orientation (is an even or direct motion)

can be written as the composition of reflections about two intersecting lines

41 Glide Reflections

Composition of three reflections produces a glide reflection.

R3(line3)° R2(line2)° R1(line1) = GR(distance, line)

Glide reflections are “opposite” motions; they do not preserve orientation; they reverse it. They are also called “odd”

Let’s look at a glide reflection:

GR(3cm,L)(ABC)

This means take the original set ABC and reflect it about line L, then translate the interim image 3 cm parallel to L and downward. Notice that the “glide” is a translation parallel to the line of reflection. Since the direction is given by L, we need not specify an angle of elevation for the translation part.

C C'' D B'' B A'' D''

The first reflection is the “reflect” in glide reflection, the second two reflections are hidden in the “glide” part of the name and are a translation.

42 Here’s an initial set: ABCD. Let’s glide reflect it about the given line 6 cm: GR(6, L1).

m G'FG = 40.00

G' L1

D B

F G horizontal

The first thing I’m going to do is translate it along L1 6 cm. Note that to do this in Sketchpad I needed to measure the angle of elevation. This translation would use two of the lines of reflection. They will be parallel lines and perpendicular to L1. Sketch in lines that would work. remember: perpendicular to L1. The interim set is in dotted lines.

m G'FG = 40.00

G' L1

D B

F G horizontal

The last thing to do is reflect about L1.

Or – I could have reflected first and then translated. The image set will be the same no matter what order I do it in.

m G'FG = 40.00

G' L1

D'' C

A'' B'' C'' D B

F G horizontal

Note that starting a A with the initial set and moving clockwise:

ABCD.

Starting with A’’ on the image set and moving clockwise:

A”D”C”B”

Which is NOT a permutation of the initial labeled points. So the orientation is reversed. (this is an odd motion in the plane or an odd mapping – these are alternate ways of saying that the orientation is reversed).

Does it matter if you reflect first then glide or can you glide first then reflect?

44 Let’s check:

GR (2”, L)

First reflect then glide: glide first, then reflect:

You should get the same end result. The circle is on the opposite side of the line and down 2 inches.

45 vnet: Glide Reflections: how to get the reflection lines

Let’s do a glide reflection as the composition of three reflections:

L1 L2

Initial se t L3

Now look at the instructions:

R3 (L 3 )° R 2 (L 2 ) ° R 1 (L 1 )(initial set)

Note that the reflection lines for the “glide” part of the motion are perpendicular to the reflection line for the “reflection” part. This is always true

Let’s track the orientation: reflect L1 – reversed reflect L2 – back to original reflect L3 – reversed again.

Glide reflection exercise:

46 Give the following transformation in composition notation. Put in your own reflection lines for the “glide” (ie translation) part.

m B'AB = 125.00

B' C E C'' D''

E''

A B

47 Properties of an glide reflection:

GR(distance, line)

preserves distances preserves angle measures preserves betweeness

changes location

reverses orientation (is also called “odd” or indirect)

can be written as the composition of three reflections with two lines of reflection parallel to each other and perpendicular to the third.

48 Answers to exercises Dilation exercise 1:

A A ' P

midpoint

midpoint'

BA = 6.00 c m B

B'A' = 9.00 cm midpoint'A' = 4.50 c m

A '

BA = 6.00 cm midpoint midpoint'

B B'

B'midpoint = 4.50 cm B'A' = 9.00 c m

The length of the image sets is the same under both transformations. The LOCATION of the image sets is different with different points of dilation.

49 Exercise with a right triangle, showing the creation of similar triangles

Suppose we have a 3cm – 4cm – 5cm right triangle, ABC.

5cm 3cm

A 4cm B

2 We want to use a scale factor of and as center of dilation the vertex of the right angle, 3 point A. Our instructions are 2 S( , A)(ABC). C 3 C'

Here’s a picture of what happens.

A B' B

A. What is the relationship between ABC, the original set, and AC’B’ the image set? The image set is smaller, nested inside the initial set. They are similar triangles.

B. What is the perimeter of the original triangle? 12 cm

2 C. What is the perimeter of the image triangle? (12)= 8cm 3

D. What is the area of the original triangle? ½ (3)(4) = 6 cm²

2 8 E. What is the area of the image triangle? ( )2� 6 cm 2 3 3

50 F. What is the relationship between CB and C'B' ? How do you know? The image is 2/3 the length of the inital

G. What are the measures of the angles of AB’C’? The same

H. If we put the original set in the Cartesian coordinate plane with vertex A at the origin, what are the coordinates of vertex C and vertex B? (0, 3) and (4, 0)

I. What are the geometric coordinates of the image vertices C’ and B’? The Cartesian coordinates are (0, 2) and ( 8/3, 0). The multiplier is 3 5 1+ ( - )2 = 4 4 C’ = 0 B’ = 10/3

2 J. Do these new coordinates reflect a similarity transformation of ? 3 Yes, of course.

51 Isosceles Right Triangles Exercise:

Suppose we have an isosceles right triangle with the congruent legs measuring 3 cm and we want to perform a similarity transformation using a point C that is nearby like this.

A B

S(2, C)(ARB) gives us

A B

A' B'

A. What is the length of side A’R’? B’R’? both are 6 cm

52 The corresponding sides under the transformation are parallel. B. What does this tell you about the relationship between the two triangles?

Yes they are similar, so the angle measures of corresponding angles are congruent.

If you connect the image, the original and C along each vertex, you get straight lines that look like this:

A B

A' B'

These lines are called perspective lines or projection lines.

Note that C is sometimes called “the vanishing point”. Can you guess why?

Look how tiny the triangle is as you approach C. See how the projection lines become concurrent at C. The triangle “vanishes” at C.

53 Here’s a hint: S(0.25, C)(ARS) is going to show up with vertices A’’, B’’, and R’’… look where it is located.

C k = 0.25 R''

A" B" original set R

R' k = 2 A B

A' B'

C. What would we use as a scale factor to go from the smallest triangle back to the original triangle? From the largest triangle back to the original triangle?

¾; ½

D. Are the angle measures in corresponding angles congruent for all three triangles?

yes

E. What is the area of each triangle? How are the areas related to the scale factors?

Area initial: 9/2; area smallest: 1/16(9/2); area largest (4)(9/2)

54 Dilation Exercise 2:

S(2, C) (circle )

2 Area c'2 = 50.27 cm

C C' P

c'2 c' S(2, P)(circle 1 2 Area c'1 = 50.27 cm

The image circles are exactly the same size, but in different locations. The one dilated about the center is concentric to the initial set. The one dilated about P is to the left and overlapping the initial set.

55 Permutation exercise:

Given the points 1234, list all the permutations of them.

2341, 3412, 4123.

Note that 2134 is NOT a permuation. 1 “jumped” over 2 and this is not allowed. It has to jump to the end of the queue.

Initial Exercises:

A. Is this a reflection? Why or why not?

No, the line of “reflection” isn’t the perpendicular bisector of lines drawn from initial point to image point. See the dotted line.

B. sketch in the line of reflection how do you know you’re right? It IS the perpendicular bisector of a line joining an initial point to an image point. I used D and D’ but you can use ANY point and it’s image.

C G m DFG = 90.00 B DF = 1.33 cm FD' = 1.33 cm C' D F D' B' A E

56 Reflection Exercise:

AA' = 3.00 cm A"A' = 3.00 cm m AA'B = 90.00 m A"A'B = 90.00 m C'CC" = 60.00 m C"'C'C"" = 60.00 CC' = 3.00 cm   B C'C"" = 3.00 cm C C' C" C"'

A A' A"

C' C""

Note that you really didn’t have to measure much of anything if you remember that a reflection is an isometry. You DO need to assure yourself that L is the perpendicular bisector of an initial point and the corresponding image point.

Starting a C and going clockwise on the initial set: CC’’AC’

Starting at the image of C: C’ and going clockwise around the reflected image: C’A”C”’C””…note the position of A…this is NOT a permutation of the initial points.

If it WERE orientation preserving, the initial points would transform to C”C”’A”C”” or a permutation of this order…and it’s not at all like this.

57 T(3 cm, 60°)(AB) exercise:

AA ' = 3.00 cm BB ' = 3.00 cm

m A'AC = 60.00

A C

B D m B'BD = 60.00

The angle of elevation off the horizontal is shown for two pairs of (initial, image) points. The distance from initial to image is per the instructions. And note that the segments connecting initial to image are parallel lines.

AA ' = 3.00 cm BB ' = 3.00 cm

m A'AC = 60.00

A C

B D

Create a diameter for circle AB that is 60° up from the horizontal. I picked the point where this diameter meets AB on the far side of the circle. I put in a reflection line that is perpendicular to the diameter and called it L1. I reflected AB about L1 (and didn’t name the interim reflection – it is the circle below L1).

Next I located the center of the interim image. I calculated the distance from this interim center to the initial center and I added 3 cm. I divided this sum in half and put in the midpoint of the distance from the interim center to the desired final center.

58 AA ' = 3.00 cm BB ' = 3.00 cm

m idpoi nt m A'AC = 60.00

A C

B D

This point is called “midpoint”. I ran a line through midpoint that is perpendicular to the line that goes through the circle’s centers at 60°…in other words it is PARALLEL to L1. This is my second line of reflection. I reflected the interim un-named circle about this line and landed right on top of the image circle…which is where I wanted to be.

Note that these are NOT the only lines of reflection that could be used. It is a good idea to use reflection lines that are perpendicular to the angle of elevation line. And, inevitably, you often end up doing some sort of calculation of where you are and where you want to be and putting the second reflection line at that halfway mark.

T(6 cm,45)(ABCD) exercise:

m ECC' = 45.00 CC' = 6.00 cm B'

C E A'

B CC' i s parall el to AA '

A F m A'AF = 44.81 AA ' = 6.00 cm

59 To set up lines of reflection:

_interim image

_C'

_D'

_B'

_A'

_D _L1 line_ att 45_ to horizontal

I put in a horizontal from A then put in a line at 45° to the horizontal. And then I put a perpendicular to the second line and called this third line L1. I then reflected the initial set about L1.,,getting the flag that is done in dotted lines.

Now, then, I want a line of reflection between the interim image and the final image… where does the reflection line go? Remember it’s the perpendicular bisector of the line joining any point of the interim set and the final set. Sketch it in first and then look below to see if you’re right.

60 DP = 6.79 cm i nterim im age m PDH = 45.00 C'

P D' L1 i s parall el to L2 L2

D H L1 l ine at 45  to hori zontal

We would write:

R(L2 )° R(L 1 )(ABCD) and this will give the same image set at the instructions written in translation notation.

61 Translation exercise:

A''

B''

B B'''

The distance is about 10 cm and the angle of elevation is about 45° so you should write:

T( 10, 45°)(BB’’’)

62 Distinguishing exercise:

The first one is a reflection; the second is a rotation.

Rotation exercise:

Here’s one answer:

FA is L2 and AB is L1. Remember the hollow dot is read “after” and you list the last reflection on the left and move right to the first reflection

R2 (FA)° R 1 (AB)(BCDE) and Rot(120° ,A)(BCDE)

F B

E' A C D

The rotation is 125 degrees

C' B' D' E

Remember you may have made different choices for the reflection lines and STILL be doing it right. This issue makes grading the homework in this section a lengthy job – each person does it right DIFFERENTLY. Oh, boy.

63 Glide reflection exercise:

R3 (L2)° R 2 (L1) ° R 1 (AB')(CDE)

B' C E C'' D'' L1 D

L2 E''

A B

Your choices of L1 and L2 may be placed differently than mine, BUT they should be parallel to each other and perpendicular to AB’.

64 65