Transformation Geometry Module

TRANSFORMATION GEOMETRY MODULE

FAJAR ARWADI

MATHEMATICS DEPARTMENT MATHEMATICS AND NATURAL SCIENCE FACULTY UNIVERSITAS NEGERI MAKASSAR 2018

QUOTES

peace be upon him) said: “Acquire – ﷺ) The Prophet Muhammad knowledge and impart it to the people.” – (Sunan Tirmidhi, Hadith 107)

“Where there is matter, there is geometry” Johannes Kepler

“Meaning is important in mathematics and geometry is an important source of that meaning”. David Hilbert

Preface

The writer aims to thank to the Almighty God, Allah, because of His bless and grace, this module titled “Transformation Geometry” can be finished. Moreover, the writer expresses gratitude to the Mathematics Department lecturers, peculiarly, geometry field lecturers.

This module is intended as a completion of geometry literatures which uses English as the medium of instruction. Therefore, it can be used a reference for International Class Program students. Related to the contents, it consists of several transformations on the Euclidean plane, i.e. reflection, halfturn, translation, and rotation. Besides that, there are some concepts related to the transformations concerned in this module namely bijective function, isometry, and the composition of two transformations.

The writer hopes this module is certainly useful for everyone, particulary for Mathematics Department students. However, critiques and advices are emphatically needed for the refinement of this module in future.

iii

LIST OF CONTENTS Page Title ...... i Quotes ...... ii Preface ...... iii List of contents ...... iv Transformation ...... 1 Reflection ...... 8 Isometry ...... 15 Composition of Transformations ...... 22 Halfturn ...... 29 Translation ...... 37 Rotation ...... 48 References ...... 56

CHAPTER I TRANSFORMATION

The discussion of transformation geometry commences with the introduction of the concept of function that have been studied in the subject of calculus. The concept underlies the topic of transformation geometry, for example, one of the postulates in Euclidean geometry, i.e. every angle on a plane is associated with exactly one real number.

The function which is discussed here is restricted to the function having domain and origin in the form of V (V is the Euclidean plane). The function definition is given as follows:

Definition :

A function of V to V is a mapping that associates every element of V to exactly one element of V

In general, a function is notated with the letter. 푓. If 푓 is a function from V to V that associates every x ∈ V to y ∈ V, it can be written as 푦 = 푓(푥) where 푥 is called the pre-image of 푦 by 푓 and 푦 is called the image of 푥 by 푓. The origin and the range of the function are 푉.

In the calculus course, the types of functions are described, however, here only three types are discussed, namely: a. Surjective Function

A function f is called a surjective function if for every y element V, there exists x element V such that 푓(푥) = 푦. And to show that a function is surjective, we must show that every element (image) has pre-image. In other words, for every element in codomain has pair in domain V. b. Injective Function

A function 푓 is called injective if for every a and b element of the domain, where a ≠ b then f(a) ≠ f(b). The statement is equivalent to “if f(a) = f(b) then a = b”. To show that a function is injective, we must show that for every pair a and b element domain, if a ≠ b then we show f(a) ≠ f(b), or if f(a) = f(b) then we show a = b.

1 c. Bijective Function

A function f is called bijective if f is surjective function and injective function. So, to show that a function is bijective, we must show that the function is both surjective function as well as injective function.

In Mathematics subject in Junior High School (SMP) and Senior High School (SMA) we have learned about symmetry, rotation, translation, and dilatation. All we have learned are the equivalent of bijective and those are transformations that will be discussed.

Whereas, transformation geometry term can be interpreted as a branch of geometry that discusses transformation, but it can also be interpreted as a geometry which is based on the transformation. The following discussion presents the geometry in the first interpretation, but at the same time it leads to the second interpretation.

Definition A transformation on plane V is a bijective function of which both the domain and the codomain are V. Where V is the Euclidean plane

To show that a mapping of V to V is a transformation, then the steps which should be consecutively undertaken are checking whether:

1. The mapping is a function. 2. The mapping is surjective. 3. The mapping is injective.

Surjective means that if ∀ B ∈ V, ∃ A ∈ V such that T(A) = B. B = mapping from A by T, and A = pre-image of B by T. T(A) = A′

Injective means that if A and B are elements of the domain, then “(A ≠ B) ⟹ (A′ ≠ B′)”, where T(A) = A’ and T(B) = B’.

The statement is equivalent to “(A′ = B′) ⟹ (A = B)”

So, to show that a function is injective, then it must be shown that for each pair of elements of the domain A and B, if A ≠ B then we must show that A′ ≠ B′, or if A′ = B′ then it must be shown that A = B.

A surjective and injective function is a bijective function. Therefore to show whether the function is bijective, it must be shown that whether the function is both surjective and injective.

Example 1:

Suppose A ∈ V. There is a mapping T of which both the domain and the codomain are V.

T: V → V is defined as follows:

1) T(A) = A. 2) If P ≠ A, then T(P) = P′ with P′ the middle point 퐴푃̅̅̅̅.

Show that the mapping T is a transformation.

Solution:

The image of point A is the point A itself.

Take a point R ≠ A on V.

Since V is the Euclidean plane, then there is one line passes through the points A and R.

It means that there is only one line segment AR, so that there is exactly one point S with S ∈ 퐴푅̅̅̅̅ , such that AS = SR.

Since R is an arbitrary point, it means that for each X ∈ V, there is Y ∈ V, where Y = T(X).

Therefore T is a function

Is T surjective?

In other words, does every point in V has a pre-image?

To answer these questions, it must be shown that for arbitrary point Y ∈ V, is there X ∈ V so that T(X) = Y.

According to the first condition, if Y = A, its pre-image is A itself, because T(A) = A. If Y ≠ A, since V is the Euclideanplane, then there is exactly one X with X ∈ 퐴푌̅̅̅̅ such that AY = YX. It means that X is pre-image of the point Y.

Thus, every point in V has a pre-image which implies that T is a surjective mapping.

Is T injective?

To show that T is injective, take any point P, Q ∈ V with P ≠ Q.

- The first case if P, Q and A are not collinear. It will be shown that the position of P′ = T(P) and Q′ = T(Q)

Suppose P′ = Q′.

Since P′ ∈ 퐴푃̅̅̅̅ and Q′ ∈ 퐴푄̅̅̅̅, then 퐴푃̅̅̅̅ and 퐴푄̅̅̅̅ has two intersection points, namely point A and point P’ or Q’. It means 퐴푃̅̅̅̅ and 퐴푄̅̅̅̅ coincide Since Q ∈ 퐴푄̅̅̅̅ and 퐴푃̅̅̅̅ = 퐴푄̅̅̅̅ then Q ∈ 퐴푃̅̅̅̅, resulting in point P, Q and A are collinear.

It is contrary to the fact that P, Q and A are not collinear. It means that the assumption that P′ = Q′ is not true, and it should be P′ ≠ Q′.

So if P ≠ Q then P′ ≠ Q′, which means injective.

- The second case is if P, Q and A are collinear (it is also injective)

From the proof above, it turns out that T is an injective mapping.

Because T is a surjective and injective mapping then T is bijective. So T is a transformation.

Example 2:

Given relation T [(x, y)] = (2x + 1, y − x).

Show that this mapping is a transformation.

Solution:

If P(x, y) then P′ = T(P) = (2x + 1, y − x). It means that the domain of T is the whole plane V.

Is T surjective?

Take an arbitrary point 퐴(x, y), is 퐴 a domain of T ?

Suppose that B(x’, y’) is a domain of point A, then certainly T(B) = A or T[(x’, y’)] = (x, y) ↔ (2x′ + 1, y′ − x′) = (x, y)

푥−1 2푦+푥−1 we get 푥′ = 푎푛푑 푦′ = . 2 2

x−1 2y±1 x−1 2y±1 So B = ( , ), hence T(( , ) = (x, y). 2 2 2 2

Since (x’, y’) always exists for each (x, y) then B (domain of A) always exists meaning that T(B) = A. Because A is an arbitrary point in V, then each point in V has domain which means that T is surjective.

Is T injective?

Take points P(x1, y1) and Q(x2, y2) where P ≠ Q

Is P’ ≠ Q’ ?

Suppose that P’ = Q’ then ( 2x1 + 1, y1 − x1) = (2x2 + 1, y2 − x2)

Since 푥1 = 푥2 and 푦1 = 푦2 then P = Q. It is contrary to the fact that P ≠ Q.

It means that the assumption that P’ = Q’ is false and it should be P’ ≠ Q’.

It is now proven that if P ≠ Q then P’ ≠ Q’. So, T is injective.

Therefore T is injective and surjective, in other words, T is a transformation

Excercises

1) Let point K and line segment of ̅퐴퐵̅̅̅ where K ∉ ̅퐴퐵̅̅̅ and a line 푔 so that 푔 // ̅퐴퐵̅̅̅ and the distance between K and ̅퐴퐵̅̅̅ is two times the distance between K and 푔. There is a mapping T with domain ̅퐴퐵̅̅̅ and the range 푔 so that if P ∈ ̅퐴퐵̅̅̅ then T(P) = P’ = 퐾푃̅̅̅̅ ∩ g. a. What is the range of P’ if P moves through ̅퐴퐵̅̅̅. b. Prove that T is injective. c. If E and F are two points on ̅퐴퐵̅̅̅, what can be interpreted about the distance of E’F’, if E’ = T(E) and F’ = T(F).

2 2 2 2 2) Let O(0,0), C1 = {(x, y)│x + y = 1}, C2 = {(x, y)│ x + y = 25}.

T: C1 → C2 is a mapping defined as: “if P ∈ C1, then T(P) = P’ OP⃗⃗⃗⃗⃗ ∩ C2 a. If A(0,1), determine T(A). b. Find the domain of B(4,3). c. If D is any point on domain T, find the distance between DD’, D’ = T(D). d. If E and F are two points on the domain of T. What can be interpreted about the distance between E′F′? 3) 퐿푒푡 퐹 ∶ 푉 −> 푉 , 𝑖푓 푃 (푥, 푦) 푡ℎ푒푛 푓(푃) = (│푥│, │푦│) a. Determine 푓(퐴) if 퐴(−,6) b. Determine all the pre-images of the point 퐵(4,2) c. What is the shape of the range? 4) Let function 푓 ∶ axis 푥 −> 푣 defined as: “if 푃(푥, 0) then 푓(푃) = (푥, 푥2) a. Find the image of 퐴(5,0) by 푓 b. Is 퐵(−13,169) ∈ image of 푓 c. Is f surjective? 5) Let a line 푠 and point of 퐴, 퐵, 퐶 as shown below

●A

●B

●C 푇: 푉 → 푉 is defined as follows: i) if 푃 ∈ 푆 then (푃) = 푃 ii) if 푃 ∉ 푆 then 푇(푃) = 푝’ such that is axis of 푃푃̅̅̅̅̅′

a. Draw 퐴’ = 푇(퐴), 퐵’ = 푇(퐵) b. Draw the pre-image of point 퐶 c. is 푇 a tranformation? d. Prove that 퐴’퐵’ = 퐴퐵 6) Let two lines 푔 and h are parallel in the Euclidean plane 푉, and a point of 퐴 which is in the middle between 푔 and ℎ. T is the image of the domain of g defined as : 𝑖푓 푃 ∈ 푔 푡ℎ푒푛 푃’ = 푇(푃) = 푃퐴̅̅̅̅ ∩ ℎ a. What is the range of T b. If 퐵 ∈ 푔, 퐶 ∈ 푔, and 퐵 ≠ 퐶, 퐵’퐶 = 퐵퐶 푤𝑖푡ℎ 퐵’ = 푇(퐵), 퐶’ = 푇(퐶) c. is 푇 injective? 7) Let three different points 퐴, 퐸, 퐷 not collinear and a relation 푇 defined as: 푇(퐴) = 퐴, 푇(푃) = 푃’, such that P is the middle point of 퐴푃̅̅̅̅. a. Draw 퐸’ = 푇(퐸) b. Draw 푄 so that 푇(푄) = 퐷 c. is 푇 a transformation?

CHAPTER II REFLECTION In senior high school physics, we have learned about the properties of reflection. It says that if an object is x units in the front of a mirror, then the image of the object is also located as far as x units behind the mirror, and if the object is located on the mirror, its image will coincide with the object. It is further discussed geometrically in this reflection section:

Definition: Reflection in line 푠 is a function 푀푠 that is defined for each point P on plane V as follows:

i. If P ∈ s so Ms = P. ̅̅̅̅̅ ii. If P ∉ s so Ms (P) = P’ , in a way such that s is the axis of 푃푃′.

Reflection in line s is notated Ms. Line s is called the axis of reflection or mirror line. As shown in the previous section, to show whether reflection is a transformation, it must be shown whether the reflection is a function that is surjective and injective. To show that a reflection is a transformation, the steps which should be undertaken are answering the following questions: 1. Is reflection a function? Based on its definition, reflection is a function from V to V. 2. Is reflection surjective? Take arbitrary point A 'on V. If A’∉ s. Geometrically, A is the element of V so s become the axis 퐴퐴̅̅̅̅̅′ (since V is an Euclidean plane).

It means that the MS(A) = A’ implying that every A' has pre-image. Then, M is surjective 3. Is reflection injective? Take two arbitrary points A, B ∈ V where A ≠ B.

There are three possibilities, namely: a. A ∈ s dan B ∈ s

It means Ms(A) = A’ = A and Ms(B) = B’ = B Since A ≠ B it means A’≠ B’ Then M is injective. b. A ∈ s and B ∉ s

It means 푀푠(퐴) = 퐴’ = 퐴 and Ms(B) = B’ such that s is the axis 퐵퐵̅̅̅̅̅′. Because A ∈ s and B’ ∉ s , so that A’≠ B’ Then M is injective. c. A ∉ s and B ∉ s

Assume that Ms(A) = Ms(B) or A’ = B’. Since 퐴퐴̅̅̅̅̅′ ⊥ s and 퐵퐵̅̅̅̅̅′ ⊥ s , so 퐵퐴̅̅̅̅̅′ ⊥ s , thus it is obtained that from point A’, two distinct lines can be created perpendicular to line s which is impossible. Then the assumption that MS(A) = Ms(B) or A '= B' is false. Thus, it should be A’≠B’. Thus, if A ≠ B then A’≠ B’.

Then, M is injective.

Since MS is a function that is both surjective and injective, Ms is transformation.

Theorem:

Every reflection in a line is a transformation

Suppose that s: ax + by + c = 0 , and P (x,y), where Ms(P) = P’(x’, y’).

● P(x,y) s

●P’(x’,y’)

If P ∉ s so 푃푃̅̅̅̅̅′ ⊥ s , then:

푦′−푦 푏 = ...... (i) 푥′−푥 푎 푥+푥′ 푦+푦′ If Q is the midpoint of 푃푃̅̅̅̅̅′ , then Q( , ) lies on the line s. 2 2 푥+푥′ 푦+푦′ So a( ) + b( ) + c = 0...... (ii) 2 2 From the equation (i) and (ii), it is obtained that:

2푎(푎푥+푏푦+푐) x’ = x - 푎2+ 푏2 2푎(푎푥+푏푦+푐) y’ = y - 푎2+ 푏2 so, if s: ax + by + c = 0, and P(x,y), then Ms(P) = P’(x’,y’) where

2푎(푎푥+푏푦+푐) x’ = x - 푎2+ 푏2 2푎(푎푥+푏푦+푐) y’ = y - 푎2+ 푏2

Example 1: If on V there is an orthogonal axis system with A (1,3) and B (-2,1). Determine the equation of the line s so Ms (A) = B!. Solution:

Ms(A) = B means that s is the axis ̅퐴퐵̅̅̅. If T is the midpoint of ̅퐴퐵̅̅̅, then s passes through the point 2 3 1 T and perpendicular to ̅퐴퐵̅̅̅. m퐴퐵⃡⃗⃗⃗⃗ = => mS = - , and T (- , 2). 3 2 2 So, the line s : y – 2 = - 3 (x + 1 ) or 2 2

s : 3x + 2y - 21 = 0 or s : 6x + 4y – 5 = 0 2 Example 2 : Suppose line s : 2x – 3y + 5 = 0

a. Determine Ms(A) if A(2,-5)

b. Determine Ms(O)

Solution: Given line s : 2x – 3y + 5 = 0 => a=2 , b=-3 , and c=5

a. A(2,-5) , Ms(A) = A’(x’,y’) 2푎(푎푥+푏푦+푐) x’ = x - 푎2+ 푏2 2.2[2.2+(−3).(−5)+5] x’ = 2 - 22+ (−3)2 96 = 2 - 13 −70 = 13 2푏(푎푥+푏푦+푐) y’ = y - 푎2+ 푏2 2.(−3)[2.2+(−3).(−5)+5] y’ = -5 - 22+ (−3)2 −144 = -5 - 13 79 = 13 −70 79 Thus, Ms(A) = A’ ( , ) 13 13

c. Ms(O) = O’(x’,y’) 2푎(푎푥+푏푦+푐) x’ = x - 푎2+ 푏2

2.2[2.0+(−3).0+5] = 0 - 22+ (−3)2 20 = 0 - 13 −20 = 13 2푏(푎푥+푏푦+푐) y’ = y - 푎2+ 푏2 2.(−3)[2.0+(−3).0+5] = 0 - 22+ (−3)2 (−30) = 0 - 13 = 30 13 −20 30 Thus, Ms(O) = O’ ( , ) 13 13

Exercises : 1. Suppose line g = {(x,y)|y = x}

a. If A(2,-3), determine Mg(A).

b. If B’(-3,5), determine pre-image of B’ by Mg.

c. If P(x,y) any point, determine Mg(P). 2. Suppose h = {(x,y)|y = 2} a. If C(3,√2 ), determine C’.

b. If D’(2,-4), determine pre-image D, by Mh. c. If P(x,y), determine P’. 3. Suppose s = {(x,y)|x = -3}

a. If A(4,1), determine A’ = Ms(A).

b. Determine the coordinate of point C if Ms(C) = (-2,7).

c. If P(x,y) is any point, determine Ms(P). 4. Suppose line l = {(x,y)|2x + 3y = 11}

a. Determine Ml(O).

b. Determine Ml(E) with E(1,2).

c. If F(x, 2x-1), determine the coordinate F if Ml(F) = F.

5. Suppose line s = {(x,y)|2x + y = 1} and t = {(x,y)|x = -2}. Find the equation of line s’ = Mt(s). 6. Suppose line t, circle l with center D, and ∆ ABC as shown below :

l ● D B t

a. Draw Mt(∆ ABC)

C b. Draw Mt(l) 7. If lines g = {(x,y)|y = 1}, h ={(x,y)|y = x} and k ={(x,y)|x = 3}. Find the equation of the following lines :

a. Mg(h) c. Mh(g)

b. Mg(k) d. Mh(k)

8. Mk is a reflection that connects point A(4,8) to point B(8,0). Determine the equation (image) of circle (x + 1)2 + (y – 3)2 = 9, if it is reflected to the line k.

9. If two points P and Q. Draw a line t so that Mt(P) = Q and determine Mt(Q). 10. There are an orthogonal axis system in V, point A(1,3), and point B(-2,-1). Determine the

equation of a line of g so that Mg(A) = B. 11. Given two parallel lines g and h, points A and B as shown in the figure. Draw the shortest path from A to B providing that it must be reflected on g then on h. g

● B

A● h

12. If g = {(x,y)|y = -x} and h ={(x,y)|3y = x + 3}

Show that whether point A(-2,-4) lies on line h’ = Mg (h). 13. If line g ={(x,y)|6x – 3y + 1=0} and a point A(k,2).

Find the value of k if Mg(A) = A. 14. Two walls form an angle as shown in the figure of which it is formed by line k and line l. A ball is located in the point A. Sketch where the ball should be directed such that if it is reflected on k and on l, it will be bounced back to A.

A●

l 15. Given line g = {(x,y)|3x – y + 4 = 0} and h = {(x,y)|2x + 3y = 6}.

Determine the equation of line g’ = Mh(g), and h’ = Mg(h).

CHAPTER III ISOMETRY In daily life, many events or movements are transformations such as the movement of a table and the opening or closing of a door. The movement of a table from one place to another and the opening or closing of door do not alter the length and the width of table or doors except the position of the table or the door. Such kind of transformation is called isometry.

Definition A transformation 푇 is an isometry if for every pair of points 푃, 푄 satisfies 푃′푄′ = 푃푄, where 푃′ = 푇(푃) and 푄′ = 푇(푄).

′ If 퐴 = 푀푠(퐴), 퐵’ = 푀푠(퐵), then 퐴’퐵’ = 퐴퐵. (prove!)

Theorem Any reflection on a line is an isometry.

As being previously discussed, the result of a reflection is preserving the lenght of segment or the distance between two points, thus reflection is isometry. Besides preserving the distance between two points, isometry also has the properties as follows:

a. Mapping a line into a line Suppose g is a line and 푇 is an isometry, it will be proved that 푔′ = 푇(푔) is a line. Put the points 퐴 and 퐵 on the line 푔 (퐴 ∈ 푔 and 퐵 ∈ 푔). Suppose 푇(퐴) = 퐴′ and 푇(퐵) = 퐵′, create a line h through point 퐴′and 퐵′. It will be proved that the line ℎ = 푔′

Take an arbitrary point 푃 on 푔 such that it forms 퐴푃퐵, and let 푃′ = 푇(푃). On the line 푔, 퐴푃 + 푃퐵 = 퐴퐵. Because 푇 is an isometry then 퐴′퐵′ = 퐴퐵, 퐴′푃′ = 퐴푃, 푃′퐵′ = 푃퐵. Suppose 푃’ lies on outside of ℎ, then at ∆퐴′푃′퐵′ must be satisfied that 퐴′푃′ + 푃′퐵′ > 퐴′퐵′ since 퐴′퐵 ′ = 퐴퐵, 퐴′푃′ = 퐴푃, 퐴푃 + 푃퐵 > 퐴퐵

It contradicts to the assumption that 퐴푃 + 푃퐵 = 퐴퐵. It means that the assumption that 푃′ lies on outside of the h is not true. Suppose 푃′ lies on ℎ or 퐴′푃′퐵′. So 푔’ ⊂ ℎ The reverse direction is proved in the same way by assuming that Q' is any point on h.

Since 푇 is a transformation, it means that it satisfies the surjective properties, then there is a 푄 such that 푇(푄) = 푄′. Suppose 푄 lies outside of 푔. Using the triangle inequality, it can be proved that 푄 should be on g, so that 푄′ = 푇(푄) must be on 푔′ = 푇 (푔). Thus, it means that ℎ′ ⊂ 푔′. Since 푔′ ⊂ ℎ and ℎ ⊂ 푔′ then ℎ = 푔′ b. Preserving the size of the angle between two lines Take the three points 퐴, 퐵, and 퐶 which are not collinear. 퐴′ = 푇(퐴), 퐵′ = 푇(퐵) and 퐶′ = 푇 (퐶)

See the 훥퐴퐵퐶. According to (a), since the AB and BC are straight lines so 퐴′퐵′ and 퐵′퐶′ are also straight lines. Because 푇 is an isometry then 퐴′퐵′ = 퐴퐵, 퐵′퐶′ = 퐵퐶, and 퐴′퐶′ = 퐴퐶. It means that 훥퐴퐵퐶 ≅ 훥퐴′퐵′퐶 ′(s.s.s), which also means that the vertices of the triangles are in the same position and the same magnitude. Thus, isometry preserves the angles.

15 c. Preserving the parallels of two lines Given Line 푔 // ℎ, 푔 ′ = 푇 (푔) and ℎ′ = 푇 (ℎ). It will be proved that 푔 ′// ℎ′

Suppose the line 푔 'intersect ℎ′ at the point 푃’, then 푃′ ∈ 푔′ and 푃′ ∈ ℎ′. Since 푇 is a transformation then for 푃′ ∈ ℎ′ there must exist 푃 so that 푇(푃) = 푃′ where 푃 ∈ 푔 and 푃 ∈ ℎ. It then implies 푔 and ℎ intersect at point 푃. It contradicts to the assumption that 푔 // ℎ implying the assumption that 푔 'intersects ℎ′ is wrong, it should be 푔′// ℎ′. One consequence of that property is that if 푔⟘ℎ then 푔′⟘ℎ ′, where T is an isometry, 푔′ = 푇(푔), and ℎ′ = 푇(ℎ). If 훥퐴퐵퐶 are reflected to the

line g, the map is 훥퐴′퐵′퐶′, or 푀𝑔(훥퐴퐵퐶) = 훥퐴′퐵′퐶′

A reflection in the line 푔 maps the 훥퐴퐵퐶 on 훥퐴′퐵′퐶′. If the 훥퐴퐵퐶 with the order of the 퐴 − 퐵 − 퐶 is opposite to the clockwise direction, then the image which is 훥퐴′퐵′퐶’ having the order 퐴′ − 퐵′ − 퐶′ is clockwise direction.

Meanwhile, the rotation about the center of rotation 푂 maps 훥푃푄푅 on 훥푃′푄′푅′. If on the 훥푃푄푅, the direction of 푃 − 푄 − 푅 is clockwise direction then its image 훥푃′푄′푅′, has also clockwise direction 푃′ − 푄′ − 푅′. Concerning the further discussion of isometry phenomenon above, we shall introduce the concept of orientation of the pair of three points which are not collinear. Suppose (퐴, 퐵, 퐶) is the pair three points which are not collinear. Then through 퐴, 퐵, and 퐶, there is exactly one circle 퐼. We can circumnavigate 퐼 started from 퐴, then in 퐵, then in 퐶 and ended up back at 퐴. If the circumferential direction is the same as the clockwise direction, it is said that the pair of three points (퐴, 퐵, 퐶) has opposite orientations to clockwise direction or negative orientation. It means that the reflection of the line g that maps 훥퐴퐵퐶 into the 훥퐴′퐵′퐶′, (퐴, 퐵, 퐶) has negative orientation, and (퐴′, 퐵 ′, 퐶′) has positive orientation. While in the rotation with the center of rotation 푂 which maps 훥푃푄푅 into the 훥푃′푄′푅′, (푃, 푄, 푅) has a positive orientation and (푃′, 푄 ′, 푅′)’s orientation is also positive. Therefore

Definition

i. A transformation 푇 maintains an orientation if for any three points which

are not collinear (푃, 푄, 푅) have the same orientation as the orientation of

the triple points (푃′, 푄′, 푅′) ii. A transformation T reverses an orientation if the orientation of any three points which are not collinear (푃, 푄, 푅) is not the same as the orientation of (푃′, 푄′, 푅′) where 푃′ = 푇(푃), 푄′ = 푇(푄), and 푅′ = 푇(푅).

Moreover, we can classify isometry into direct isometry and indirect isometry (opponent isometry) by looking at the following definition:

Definition A transformation refers to direct isometry if the transformation preserves the orientation, and it is called opponent isometry if the transformation reverses the orientation.

An isometry refers to direct isometry if the isometry preserves orientation and called opponent isometry if the isometry changes the orientation. Thus, since the reflection changes the rotation, it means that the reflection is opponent isometry, while the rotation is a direct isometry since it maintains orientation. Furthermore, exactly one property holds, i.e. any isometry is either direct isometry only or an opponent isometry only (not both). Example 1 :

푇 is a transformation defined by 푇 (푃) = (푥 − 7, 푦 + 4) for all points 푃 (푥, 푦) ∈ 푉. Show whether 푇 is an isometry? Solution: Suppose that 푃’ = 푇(푃) = (푥 − 7, 푦 + 4) ∀ 푃(푥, 푦) ∈ 푉

Suppose also that Point 퐴(푥1, 푦1) and (푥2, 푦2) with 퐴 ≠ 퐵.

It means that 퐴’ = 푇(퐴) = ( 푥1 − 7, 푦1 + 4) and

퐵’ = 푇(퐵) = (푥2 − 7, 푦2 + 4)

2 2 퐴퐵 = √(푥2 − x1) + (푦2 − 푦1)

2 2 퐴′퐵′ = √{(푥2 − 7) − (x1 − 7)} + {(푦2 + 4) − (푦1 + 4)}

2 2 = √(푥2 − x1) + (푦2 − 푦1) Since 퐴’퐵’ = 퐴퐵, so 푇 isometry

Example 2 Suppose 푇 is a transformation that is defined for all points 푃(푥, 푦) as 푇(푃) = (− 푦, 푥). a. Is 푇 an isometry? b. If 푇 isometry, is it direct isometry or opponent isometry? Answer:

Solution: a. Given 푃’ = 푇(푃) = (푦, −푥) ∀ 푃(푥, 푦) ∈ 푉

Suppose that point 퐴(푥1, 푦1) and (푥2, 푦2) with 퐴 ≠ 퐵. 2 2 퐴퐵 = √(푥2 − x1) + (푦2 − 푦1) 2 2 퐴′퐵′ = √(푦2 − 푦1) + {(−푥2) − (−x1)} 2 2 = √(푦2 − 푦1) + (−푥2 + x1) 2 2 = √(푥2 − x1) + (푦2 − 푦1) evidently 퐴’퐵’ = 퐴퐵

Because 퐴’퐵’ = 퐴퐵, so 푇 isometry. b. To check whether 푇 is a direct isometry or opponent isometry, take any three points which are not collinear, for example 푂(0.0), 퐴(1,2) and 퐵(4.1). By using transformation 푇(푃) = (푦, −푥) ) ∀ 푃(푥, 푦) ∈ 푉 then it is obtained that 푂’ = (0,0), 퐴’ = (−2,1), and 퐵’ = (−1,4). Since the orientation (푂, 퐵, 퐶) is positive, and the orientation (푂′, 퐴′, 퐵′) is also positive, then 푇 is a direct isometry.

Exercises 1. 푇 is a transformation defined by 푇(푃) = (3푥 + 5,2 − 4푦) for all points 푃(푥, 푦) ∈ 푉. Show that T is an isometry. 2. Suppose the points 퐴(1, −1), 퐵(4,0), 퐶(−4,1) and 퐷(−2, 푝). If an isometry 푇 with 푇(퐴) = 퐶 and 푇(퐵) = 퐷, find the value of 푝. 3. Prove that the transformation with the formula: 푥′ 3/5 −4/5 푥 [ ] = [ ] [ ] is an isometry. 푦′ −4/5 −3/5 푦 4. Suppose 훥퐴퐵퐶 by isometry 푇 is mapped into 훥퐴′퐵′퐶′, prove that 훥퐴퐵퐶 ≅ 훥퐴′퐵′퐶′. 5. Given a line 푔. 푇 a function defined for each point of 푃 in the plane 푉 as follows: i. If 푃 ∈ 푔 then 푇(푃) = 푃 ii. If 푃 ∉ 푔 then 푇(푃) = 푃’ such that 푃′ is the midpoint of the line segment of orthogonal from 푃 to 푔. a. Is 푇 a transformation? b. Is 푇 an isometry? c. If there are two points 퐴 and 퐵 so that 퐴′퐵 ′ = 퐴퐵 where 퐴′ = 푇 (퐴), 퐵 ′ = 푇 (퐵), what can be interpreted about 퐴 and 퐵?

6. Suppose the 푆 line and the points 퐴, 퐴′, 퐵 and 퐶 with 퐴′ = 푀𝑔(퐴). By only using a

ruler without ascale, sketch point 퐵’ = 푀𝑔(퐵) and 퐶’ = 푀𝑔(퐶).

7. Given lines 푠, 푡, 푢; points 퐴 and 퐵 as in the following figure. 푇 is an isometry where 퐵 = 푇(퐴) and 푢 = 푇(푠). If 푡⟘푠, painting 푡′ = 푇(푡)

8. Suppose a line 푔 and and a circle 퐼. Prove that 푀𝑔(퐼) = 퐼’ with 퐼’ is also a circle.

9. Suppose lines 푔, 푔’, ℎ, ℎ’ and 푘 where 푔’ = 푀푘(푔) and ℎ’ = 푀푘(ℎ). If 푔’//ℎ’ prove that 푔//ℎ.

10. Suppose lines 푔, ℎ, and ℎ’ where ℎ’ = 푀𝑔(ℎ). Verifying the truth of the following expressions: a. If ℎ’//ℎ, then ℎ//푔 b. If ℎ’ = ℎ, then ℎ = 푔. c. If ℎ’⋂ ℎ = {퐴}, then 퐴 ∈ 푔. 11. Suppose line g and two points 퐴 and 퐵 as shown in the following figure:

a. by using an appropriate isometry, determine a point 푃 ∈ 푔 so that 퐴푃 + 푃퐵 as short as possible. b. If 푄 ∈ 푔 is different to the point 푃, prove that 퐴푄 + 푄퐵 > 퐴푃 + 푃퐵. 12. Suppose a circle 퐼 = {(푥, 푦) | (푥 − 2) 2 + (푦 − 3) 2 = 4}. 푇 is an isometry mapping point 퐴(2,3) on 퐴′(1,7). Find the equation of the set 푇(퐼). Is the map of I also circle? Why?

13. In the following figure, there are three points which are not collinear, it is 푃, 푄, 푅. 푇 and 푆 are isometry where 푃′ = 푇(푃), 푄′ = 푇(푄), 푅′ = 푇(푅), and 푃" = 푆(푃), 푄′ = 푆(푄), 푅′ = 푆(푅). What kind of isometry 푇 and 푆 are those?

14. Isometry 푇 maps point 퐴 to 푃, point 퐵 to 푄 and 퐶 to 푅. If 푇 is an opponent isometry, find (sketching) the position of point 푅.

15. Find the coordinates of the point 푃 on the 푥 axis, measure of ∠퐴푃푂 = ∠퐵푃푋, if 퐴 = (0,3) and 퐵 = (6,5)

16. Suppose line g and points 퐴, 퐵 and 퐴’ where 퐴’ = 푀𝑔(퐴), and 퐴퐵⃡⃗⃗⃗⃗ //푔. By using

only one ruler, determine the coordinate of point 퐵′ = 푀𝑔(퐵).

CHAPTER IV COMPOSITION OF TRANSFORMATIONS A. Composition Two Transformations If F and G are respectively a transformation, then the composition of the two transformations is defined as follows:

Definition Suppose F and G are two transformations where F: V → V and G: V → V, then the composition of F and G are notated G.F defined as (G.F) (P) = G [F (P)], ∀ P ∈V.

What about the composition of F and G, is it a transformation? To answer the question, whether the composition of two transformations is also a transformation, then the following steps must be solved consecutively: 1. Checking whether the composition of two transformations is functions 2. Checking whether the composition of two transformations is surjective 3. Checking whether the composition of two transformations is injective Therefore, suppose the composition of F and G is H, or H = G. F. • If F and G are functions, it is clear that H is also a function. • Is H surjective? Take arbitrary Y∈V. Is there X∈V such that H (X) = Y? Since G is transformation, for every Y∈V there is Z∈V such that G (Z) = Y. Similarly, since F is transformation then for every Z∈V, there is X∈V that F (X) = Z. From 퐺(푍) = 푌, it is obtained that 퐺[퐹(푋)] = 푌 or (퐺 ∙ 퐹)(푋) = 푌, so 퐻(푋) = 푌 meaning that 퐻 is surjective. • Is 퐻 injective? Take arbitrary 푃, 푄 elements of 푉 where 푃 ≠ 푄 Suppose (푃) = 퐻(푄) , then 퐺[퐹(푃)] = 퐺[퐹(푄)] Since 퐺 is injective then 퐹(푃) = 퐹(푄), and since 퐹 injective then 푃 = 푄. It is contrary to which it is known that 푃 ≠ 푄. It means that the assumption that 퐻(푃) = 퐻(푄) is not true. Therefore, it should be 퐻(푃) ≠ 퐻(푄) implying 퐻 is injective. Since 퐻 is surjective as well as injective, then 퐻 is bijective. Thus, 퐻 is a transformation.

Theorem The composition of two transformations is a transformation.

Example 1 Suppose that transformation 푇1[(푥, 푦)] = (푥 + 2, 푦) and 푇2[(푥, 2푦). If 푇 is the composition of

푇1 and 푇2, find 푇. Solution:

푇 is the transformation 푇1 and 푇2 then :

푇[(푥, 푦)] = (푇1 ∙ 푇2 )(푥, 푦)

= 푇2[푇1 (푥, 푦)]

= 푇2 [(푥 + 2, −푦)] = (푥 + 2, −2푦) So the transformation 푇 is [(푥, 푦)] = (푥 + 2, −2푦) .

The Properties of The Composition of The Transformation

It has been discussed previously that the composition of two transformations is a transformation. It means that the composition of transformations is closed. Furthermore, the composition of transformations is also associative, but not commutative.

Theorem

If T1, T2, and T3 are the transformation then T1[T2.T3]=[T1.T2].T3

B. Inverse Transformation

′ 2 If Mg(P) = P , then Mg. Mg(P) = P or Mg (P) = P.

So, M2 is a transformation that describes each point onto itself. This transformation is called the identity transformation is symbolized by the letter I.

So that, I(P) = P, ∀P.

If T is a transformation, then T. I(P) = T[I(P)] = T(P), ∀P.

So, T. I = I and I. T(P) = I[T(P)] = T(P), ∀P.

Thus I. T = T. Implied T. I = I. T = T

So that,the identity transformation I is the number 1 in the set of transformations with multiplication operation among these transformations. In the set of real numbers, by multiplication operations, each transformation T has inverse Q such that T.Q = I = Q.T.

If there is an invers transformation of T, then the transformation of T is written T-1,so that T.T- 1=T-1 .T = I

Theorem Each transformation 푇 has an inverse

Proof: Suppose 푇 is a transformation. We define the equivalent 퐹 as follows: Suppose 푋 is an element of 푉, where 푉 is an euclidean field. Since 푇 is a transformation, then 푇 is bijective implying that A∈ 푉, such that 푇(퐴) = 푋.

We define then 퐹(푋) = 퐴. meaning that 퐹(푋) is the pre-image of 푋, such that from 푇(퐴) = 푋, 푇[퐹(푋)] = 푋 or (푇. 퐹)(푋) = 퐼(푋), for every 푋 element of 푉. So 퐹푇 = 퐼, so that 푇퐹 = 퐹푇 = 퐼.

Now it shall be proved that L is a transformation. From this definition, it is clear that F is surjective. Suppose 퐹(푋1) = 퐹(푋2) and suppose 푇(퐴1) = 푋1, 푇(퐴2) = 푋2 where 퐹(푋1) = 퐴1 and 퐹(푋2) = 퐴2

Since T is a transformation, then 퐴1 = 퐴2, and it is also obtained that 푋1 = 푋2 such that F is injective. Thus, it proves that 퐹 is injective. So 퐹 is a transformation.

The transformation 퐹 is called the inverse transformation of 푇 and it is denoted by 퐹 = 푇−1.

Theorem Transformation has exactly one inverse.

Proof:

Suppose T is a transformation with two inverses, namely S1 and S2 then

(푇 ∙ 푆1 )(푃) = (푆1 ∙ 푇)(푃) = 퐼(푃) for all 푃 and

(푇 ∙ 푆2 )(푃) = (푆2 ∙ 푇)(푃) = 퐼(푃) for all 푃 so, (푇 ∙ 푆1 )(푃) = (푇 ∙ 푆2 )(푃) ⇒ 푇 [푆1 (푃)] = 푇 [푆2 (푃)] because 푇 is injective , then 푆1 (푃) = 푆2 (푃) for all 푃 Theorem Inverse of any reflection in a line is a reflection itself.

So, 푆1 = 푆2 = 푆 Proof:

A reflection in the line 푔 is Mg

For X 휖 푔 then Mg (X) = X, So Mg . Mg (X) = Mg (X) = X = I(X)

So, Mg . Mg = I Thus , the Mg−1 = 푀푔

For X ∉ 푔 , Mg (X) = X’, so the 푔 axis 푋푋̅̅̅̅ , then Mg . Mg (X) = Mg (X’) = X = I(X) -1 with 푔 axis 푋푋̅̅̅̅ . So, Mg . Mg = I atau Mg = Mg

Definition A transformation which has inverse which is the transformation itself is called an involution.

Theorem If 푇 and 푆 are transformations, then (푇 ∘ 푆) = 푆−1 ∘ 푇−1

Proof:

Since (T.S) -1 is the inverse of (T.S) then (T.S) -1.(T.S) = I. Meanwhile (S-1 .T-1) (T.S) = S-1.T- 1.T.S = S-1.I.S = S-1.S = I

-1 -1 -1 Since every transformation has only one inverse, then (TOS) = S T

Therefore the inverse of the composition transformations is the composition of the inverses of each transformations in reverse order.

Example 1

In an orthogonal axis XOY system, transformations F and G are defined as follows

For ∀ P (x, y), F (P) = (x + 2.1 / 2 y) and G (P) = (x-2,2y). (FG) (P) = F [G (P)] = F [(x-2,2y)] = (x, y) = P

While (GF) (P) = G [F (P)] = G [(x + 2.1 / 2 y) = (x, y) = P. So (FG) (P) = (GF) (P ) = P = I (P), ∀ P or FG = GF = I

Thus F and G are the mutual transformations of each other denoted by G = F-1 or F = G-1

Example 2

On an orthogonal axis system, the line g = {f (x, y) | y = x} and h = {(x, y) y = 0}

Find P such that (Mh.Mg) (P) = R with R (2,7)

Solution:

Let P (x, y)

(Mh.Mg) (P) = R => Mh (Mh.Mg) (P) = Mh (R)

→ (Mh Mh Mg) (P) = Mh(R)

→ (Mh Mh ) (Mg) (P) = Mh(R)

→ (Mg) (P) = Mh(R)

→ Mg.Mg (P) = Mg.Mh(R)

→ P = Mg.Mh(R)

implying that P (x,y) = Mg.Mh(2,7) = Mg(2,-7) = (-7,2).

So the coordinate of the point P is (-7,2)

Exercises

1. Given lines g and h. A point K is the intersection of g and h, as well as the points P and Q points on g and h. Sketch: •P g

h •Q

a. A = Mg [Mh(P)]

b. B = Mh [Mg (P)]

c. C = Mh [Mh (P)]

d. D = Mg [Mh (K)]

e. R so Mh [Mg (R)] = Q

f. Do Mg.Mh = Mh.Mg ? Why? 2. Suppose that T and S are isometries, check whether the following statements below are true and give the reason. a. TS is an isometry b. TS = ST c. If g is a line, then g’ = (TS) (g) is also a line. d. If g//h, and g’ = (TS) (g), h’=(TS) (h), then g’//h’

3. Given two intersecting lines g and h, skecth

a. K such that Mg [Mb (k)] = g

b. b m such that Mh [Mg (m)] = g

c. N such that Mh [Mg (n)] line divides the acute angle between g and h. 4. The line g is the x-axis of an orthogonal axis system and h {(x, y | y = x}. Define:

a. The line equation Mh [Mg (g)]

b. P '' = Mh [Mg (P)], with P (0,3)

c. Q’’= Mg [Mb (Q)], with Q (3,-1)

d. R’’= Mg. Mh (R) with R (x,y) e. The magnitude of

b. If P (x, y) any point, determine the coordinate of point S.Mg (P)

c. Check whether S. Mg = Mg.S.

d. Check whether S. Mh = Mh.S. 6. If g = (x, y) | y = 0} and h {(x, y) | y = x} and S is a transformation defined as question 5, whereas A (2, -8) and P (x, y) , Determine the coordinates of the following points

a. Mg Mh S(A).

b. Mg.S .Mh (A).

27 c. S Mg.S. (A). d. Mh.S .Mg (P). 2 e. S .Mh (P). 2 f. S Mg (P) 7. Suppose that g and h are two lines which are perpendicular to each other. A, B, and C are three points so that Mg (A) = B and Mh (A) = C. Determine the following points. 3 a. Mg (A) b. Mh Mg Mh (A) c. Mh Mg Mh Mh Mg (A) 2 3 d. Mg Mh (A)

8. Simplify.! -1 a. (WgVhMg) -1 b. (Mh Vh Wg Mg) 1 9. Suppose the transformations T1 [(x,y)] = (-x,y) and T2 [(x,y)]= (x, y). Find the formula for 2

T2. T1 then if T 1= T2. T1, find the equation T (g) if g = {(x, y) | x + y = 0}.

What is T2. T1 = T1. T2?

10. If two different lines g and h intersect at point P, prove that Mg Mh (A) = P if and only if A = P 11. It is known that g // h and points P, Q are neither on g nor on h. a. Sketch P’’ = Mg Mh(P) and Q’’= Mg Mh(P) b. What is the form of quadrilateral PP’’Q’’Q ? c. Prove your opinion!

CHAPTER V HALFTURN

Halfturn is a special case of rotation, where the rotation angle is 180°. Since, halfturn has special characteristic, it is discussed earlier. In the previous section, it has been discussed that a reflection is an involution. Another example of an involution is a halfturn surrounding a point. One halfturn reflects every point in a plane figure at a certain point.

Therefore, halfturn is called a reflection about point, and that point is the center of the halfturn.

Definition

Halfturn about a point 퐴 is a mapping 푆퐴 that is defined for each point 푃 on a plane as follows : i. If 푃 = 퐴 , then 푆퐴(푃) = 푃. ̅̅̅̅̅ ii. If 푃 ≠ 퐴 , then푆퐴(푃) = 푃′, where 퐴 as the center point of 푃푃′

Since halfturn is also a reflection of a point, and reflection is a transformation, then it can be said that halfturn is a transformation.

Theorem Halfturn is a transformation.

′ ′ Suppose 퐴(푎, 푏),푆퐴 map point 푃(푥, 푦) to 푃′(푥 , 푦 ), then푆퐴(푃) = 푃′ where 퐴 is the center point 푥+푥′ 푦+푦′ of 푃푃̅̅̅̅̅′ so = 푎 and = 푏. 2 2

It is obtained that 푥′ = −푥 + 2푎 and 푦′ = −푦 + 2푏.

Thus, if 퐴 = (푎, 푏)and 푃 (푥, 푦), then 푆퐴(푃) = (2푎 − 푥, 2푏 − 푦).

Theorem

If 푔 and ℎ intersect and are perpendicular at point 퐴, then 푆퐴 = 푀𝑔푀ℎ.

Proof :

Since 푔 ⊥ ℎ, we can make a system of orthogonal axis where 푔 as axis 푥 and ℎ as axis 푦, and 퐴 is used as the point of origin. It must be proved that each point 푃 (푥, 푦), satisfies 푆퐴(푃) =

푀𝑔푀ℎ(푃).

Let 푃 (푥, 푦) ≠ 퐴and 푆퐴(푃) = 푃′′(푥1, 푦1).

푥 +푥 푦 +푦 Since 퐴 is the center point 푃푃̅̅̅̅′′̅, so (0,0) = ( 1 , 1 ), so 푥 = −푥 and 푦 = −푦, therefore, 2 2 1 1

푆퐴(푃) = (−푥, −푦).

While (푀𝑔푀ℎ)(푃) = 푀𝑔[푀ℎ(푃)] = 푀𝑔[(−푥, 푦)] = (−푥, −푦)

It turns out that 푆퐴(푃) = (푀𝑔푀ℎ)(푃) = (푥, 푦)

Thus, 푆퐴 = 푀𝑔푀ℎ

Theorem

If 푔 and ℎ are two lines perpendicular to each other, then 푀𝑔푀ℎ = 푀ℎ푀𝑔

Proof

If 푃 = 퐴 (Look at the figure of the previous theorem), so 푀𝑔푀ℎ(푃) = 푀𝑔(푃) = 푃. Whereas

푀ℎ푀𝑔(푃) = 푀ℎ(푃) = 푃, so 푀𝑔푀ℎ(푃) = 푀ℎ푀𝑔(푃). if 푃 ≠ 퐴, then 푀𝑔푀ℎ = 푆퐴, while

푀ℎ푀𝑔(푃) = 푀ℎ[(푥, 푦)] = (−푥, −푦) = 푆퐴(푃)

It turns out that 푀𝑔푀ℎ = 푀ℎ푀𝑔 = 푆퐴

So,푀𝑔푀ℎ = 푀ℎ푀𝑔.

Theorem −1 If 푆퐴 is halfturn, then 푆퐴 = 푆퐴.

Proof: Suppose that 푔 and ℎ are two lines perpendicular to each other and intersect at point 퐴,

−1 −1 −1 −1 then 푀𝑔푀ℎ = 푆퐴, implying that 푆퐴 = (푀𝑔푀ℎ) = 푀ℎ 푀𝑔 = 푀ℎ푀𝑔 = 푀𝑔푀ℎ = 푆퐴. So −1 푆퐴 = 푆퐴. ′ Reflection in line 푔 which is defined as 푀𝑔(푃) = 푃 when 푃 ∈ 푔, and 푀𝑔(푃) = 푃 . with 푔 is the axis 푃푃̅̅̅̅ when the 푃 ∉ 푔. When we look in general, so for every 푃 ∈ 푔 the image of point 푃 is the point itself. Such point is called invariant point of the reflection.

Definition: Point 퐴 is called the invariant transformation of T, if it is satisfied that 푇(퐴) = 퐴

It can be seen that reflection has many invariant points that are infinite, meanwhile halfturn just has one invariant point, i.e. the center of the halfturn. It has been discussed in the previous section that isometry is a transformation which maps a line into a line. When a line by a transformation has image in the form of a line, such transformation is called a collineation. Based on the above understanding, any isometry is a collineation. Since halfturn is an isometry, it is also a collineation. Among collineations, one of them is dilatation, defined as follows :

Definition:

A collineation (∆) is called dilatation, if for every line 푔, it satisfies the property

∆푔//푔. One example of collineation which is dilatation is halfturn. The example was verified by the following theorem :

Theorem:

Suppose 푆퐴 is a halfturn and 푔 is a line, if 퐴 ∉ 푔, then 푆퐴(푔)//푔

Proof :

′ Suppose 푃 ∈ 푔 then 퐴 is the midpoint of the segment ̅푃푃̅̅̅̅’ with 푃 = 푆퐴(푃). ̅̅̅̅̅ ′ Suppose 푄 ∈ 푔 then 퐴is the midpoint of the segment 푄푄′ with 푄 = 푆퐴(푄).

Since ∆퐴푃푄 ≅ ∆퐴푃′푄′ then 푃푄푃′푄′ a parallelogram, implying that 푃푄 // 푃′푄". Thus, 푔 // 푆퐴(푔).

Example 1 : Suppose that two lines 푔 and ℎ are not parallel, 퐴 is a point not located neither on 푔 nor ℎ. Determine all points 푋 on 푔 and all points 푌 on ℎ such that 퐴 is the midpoint of the segment 푋푌̅̅̅̅.

Solution :

′ ′ Take a point 푃 ∈ 푔. Sketching 푃 = 푆퐴(푃). Then 푔 = 푆퐴(푔) passes through 푃′ where 푃퐴 = 퐴푃′, 푔′ // 푔. If 푔′ intersecting ℎ in 푌, then draw a line YA intersecting 푔 in 푋. Then 푋 and 푌 are the pair of the points which seems to be exactly the only one pair a. Prove that 푋 and 푌 is the only pair that satisfy the condition. ′ ′′ b. If we didn’t use 푔 = 푆퐴(푔) but we used ℎ = 푆퐴(ℎ), could we get another pair?

Theorem :

A halfturn is a dilatation which has involutoric characteristic. Proof : Suppose 푃 is the center of a halfturn, and 푔 is a line. We must prove that: a. 푆푃(푔) // 푔 b. 푆푃푆푃 = 퐼, where 퐼 is an identity transformation

Prove: a. It means that 푆푃(푔) = 푔′ is a line. Suppose 퐴 ∈ 푔, 퐵 ∈ 푔, then 퐴′ ∈ 푔′, and 푃퐴 = 푃퐴′, 푃퐵 = 푃퐵′, while 푚 (∠퐴푃퐵) = 푚(∠퐴′푃′퐵′). Since ⊿ 푃퐴퐵 ≅ ⊿ 푃퐴′퐵′, 퐴퐵퐴′퐵′ is a parallelogram implying 푔′// 푔. ′ b. Since 푆푃푆푃(퐴) = 푆푃(퐴 ) = 퐴 for all points 퐴 ∈ 푔, then 푆푃푆푃(푔) = 퐼(푔).

So 푆푃푆푃 = 퐼 meaning that 푆푃 is an involution.

The Composition of halfturns

The characteristics of the composition of halfturns are classified according to their centers and whether there is an invariant point.

Theorem The composition of two halfturns with different centers does not have invariant point.

Proof: Suppose 퐴 and 퐵 are the centers of the halfturns. Suppose 푔 = 퐴퐵, both ℎ and 푘 are perpendicular line to 퐴퐵 in 퐴 and 퐵 respectively, then:

푆퐴푆퐵 = (푀ℎ푀𝑔)(푀𝑔푀푘)

= [(푀ℎ푀𝑔)푀𝑔]푀푘

= [푀ℎ (푀𝑔푀𝑔)]푀푘

= 푀ℎ 퐼 푀푘

= 푀ℎ푀푘

Suppose that 푋 is the invariant point of SASB, then 푆퐴푆퐵(푋) = 푋 implying (푀ℎ푀푘) (푋) = 푋 or

푀푘(푋) = 푀ℎ(푋).

Suppose 푀푘(푋) = 푋1

If 푋 ≠ 푋1, then each ℎ and 푘 is 푋푋1 axis and since a line segment has exactly one axis, it must be ℎ = 푘. It is certainly not possible since 퐴 ≠ 퐵.

If 푋 = 푋1 then 푀푘(푋) = 푋1 and 푀ℎ(푋) = 푋1 , So 푋 ∈ ℎ and 푋 ∈ 푘 meaning that 푘 and ℎ intersect at point 푋 which is not possible because ℎ // 푘.

Indeed, there can not be a point 푋 such that 푀퐾(푋) = 푀ℎ(푋) or 푆퐴푆퐵 = 푋.

So 푆퐴푆퐵 doesn’t have a fixed point.

Theorem If 퐴 and 퐵 are two different points, then there is only one and a halfturn that maps 퐴 to 퐵.

Proof:

Suppose there are two halfturns 푆퐷 and 푆퐸, so 푆퐷(퐴) = 퐵 and 푆퐸 (퐵) = 퐴.

So, 푆퐷(퐴) = 푆퐸 (퐵), then 푆퐷[푆퐷(퐴)] = 푆퐷[푆퐸(퐴)], then 퐴 = 푆퐷푆퐸(퐴). If 퐷 and 퐸 are two different points, it means that 퐴 is a fixed point 푆퐷푆퐸. 퐼 which is not possible. Therefore, there is no more than one halfturns that maps 퐴 to 퐵. The only halfturn is 푆푇(퐴) = 퐵 where 푇 is the midpoint of ̅퐴퐵̅̅̅.

Example:

Given 퐸 = {(푥, 푦)|푥2 + 4푦2 = 16}, 퐴(4, −3), and 퐵(3,1).If 푔 is the 푋 axis, show that 퐴 ∈

푀𝑔푆퐵(퐸)?

Solution: −1 −1 −1 It is known that (푀𝑔S퐵) = S퐵 . M𝑔 = S퐵푀𝑔

If 푃(푥, 푦), then 푀𝑔(푃) = (푥, −푦) then S퐵(푃) = (2.3 − 푥, 2.1 − 푦) = (6 − 푥, 2 − 푦)

퐴 ∈ 푀𝑔푆퐵(퐸) ⇔ 푆퐵푀𝑔(퐴) ∈ 퐸

푆퐵푀𝑔(퐴) = 푆퐵[푀𝑔(4, −3)] = 푆퐵(4,3) = (2, −1) −1 Since (2, −1) ∉ 퐸, then (푀𝑔푆퐵) (퐴) ∉ (퐸) In a similar way, we can define a set of maps if the equations are known. −1 In the latest example we know that 푃 ∈ 푀𝑔푆퐵(퐸) if and only if (푀𝑔푆퐵) (푃) ∈ (퐸). −1 −1 If 푃(푥, 푦) so (푀𝑔푆퐵) (푃) = (6 − 푥, 2 + 푦), then (푀𝑔푆퐵) (푃) ∈ (퐸) if and only if (6 − 푥, 2 + 푦) ∈ {(푥, 푦)|푥2 + 4푦2 = 16}.

2 2 So it must be (6 − 푥) + 4(2 + 푦) = 16. 푃(푥, 푦) ∈ 푀𝑔푆퐵(퐸) if and only if 푃(푥, 푦) ∈ {(푥, 푦)|푥2 + 4푦2 − 12푥 + 16푦 + 36 = 0} Therefore 푥2 + 4푦2 − 12푥 + 16푦 + 36 = 0 is the equation of the image of 퐸 by transformation of 푀𝑔푆퐵.

Exercises 1. Suppose three distinct points A, B, P are not collinear, sketch

a. 푆퐴(푃)

b. 푆퐴푆퐵(푃)

c. R so 푆퐵(푅) = 푃 2 d. 푆퐴 (푃) 2. Given the line g and point A, 퐴 ∉ 푔 ′ a. Draw lines 푔 = 푆퐴(푔), why 푆퐴(푔) is a line? b. Prove that 푔′// 푔 3. Given ∆ABC and a parallelogram WXYZ. There is a point K which lies outside the triangle ∆ABC and the parallelogram WXYZ.

a. Sketh 푆K(∆ABC)!

b. Find a point J so SJ(WXYZ) = WXYZ 4. If A = (2,3) determine!

a. S A (C) if C (2,3) b. S A (D) if D (-2,7)

-1 c. S A (E) if E (4, -1) d. S A (P) if P (x, y) 5. If C = (-4.3) and g = {(x, y) | y = x}, determine!

a. M g S c (2.4) b. M g S c (P) if P = (x, y)

-1 c. (M g S c) (P) d. Is M g S c = S c M g? Explain 6. Give the implication from the following expressions : a. S A (k) = S A (j) b. S A (D) = S B (D)

c. S A (E) = E d. g is a line and S A (g) = g

e. If A ≠ B and S A S B // g

7. Given A = (0,0) and B = (-4.1). Determine K such that the S A S B (K) = (6,2) 8. Given A = (-1,4), g = {(x, y) | y = 2x-1} and h = {(x, y) | y = -4x}

a. Determine the set equation SA (g) = g '.

b. Determine the set equation SA (h) = h '.

c. Determine the set equation SA (axis-x)

d. Does the point (-5.6) lie on S A (g)? Explain! 9. Given circle C = {(x, y) | x 2 + (Y-3) 2 = 4}, a line g = {(x, y) | y = x} and A = (3,2).

Show whether the point D = (2,5) is the element of the set M g S A (C). 10. Given line g, point P, and circle C, and suppose P ≠ g does not intersect the C and P ∉ C. Moreover, C circle centered at A

a. By applying a halfturn, construct the line segment 퐴푇̅̅̅̅, so that the X ∈ C, Y ∈ g such that the P midpoint of 푋푌̅̅̅̅. b. Prove that the construction is correct.

CHAPTER VI

TRANSLATION

A. Vector We have previously learned line segment in other lectures. In this part, we will discuss about vector which is defined as follows Definition Vector is a line segment of which one of the edge is the initial point and the other edge is the terminal point.

The notation of vector - The notation of vector 퐴퐵 is 퐴퐵⃗⃗⃗⃗⃗ - The notation of vector 퐶퐷 is 퐶퐷⃗⃗⃗⃗⃗ Vector퐴퐵 and 퐶퐷 are vector of which the points 퐴 and 퐶 are the the initial points, and the points 퐵 and 퐷 are the terminal points. As the illustration, look at the following figure:

Definition

퐴퐵⃗⃗⃗⃗⃗ is equivalent to 퐶퐷⃗⃗⃗⃗⃗ that is notated as 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗ , if there is a halfturn

푆푝(퐴) = 퐷 where 푃 is the midpoint 퐵퐶⃗⃗⃗⃗⃗

The definition above can be interpreted also that 퐴퐵⃗⃗⃗⃗⃗ = 퐶⃗⃗⃗⃗퐷⃗ , if 푆푝(퐶) = 퐵 and 푃 is the midpoint of 퐴퐷⃗⃗⃗⃗⃗ , as an illustration, look at the following figure:

The equivalence of two vectors satisfies a characteristic asserted by the following theorem

Theorem ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ If there are two vectors 퐴퐵 and 퐶퐷 which are not collinear then 퐴퐵퐶퐷 is a parallelogram if and only if 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗

Proof:

Look at the latest figure:

1. Suppose that 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗ , if 푃 is the midpoint of 퐵퐶⃗⃗⃗⃗⃗ , then 푆푝(퐴) = 퐷, by the definition of equivalence. The diagonals of quadrilateral 퐴퐵퐶퐷 are bisected in 푃. Then 퐴퐵퐶퐷 is a parallelogram. 2. Suppose that 퐴퐵퐶퐷 is a parallelogram, then diagonals 퐴퐷 and 퐵퐶 are

intersected in the midpoint 푃 implying that 푆푝(퐴) = 퐷. In other words, 푃 is the midpoint of both 퐴퐷 and 퐵퐶.

So 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗

Consequence:

If 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗ , then 퐴퐵퐶퐷, 퐴퐵⃗⃗⃗⃗⃗ and 퐶퐷⃗⃗⃗⃗⃗ are either parallel or colinear

Theorem

Suppose 퐴퐵⃗⃗⃗⃗⃗ , 퐶퐷⃗⃗⃗⃗⃗ dan 퐸퐹⃗⃗⃗⃗⃗ are vectors, then the following properties apply:s

• 퐴퐵⃗⃗⃗⃗⃗ = 퐴퐵⃗⃗⃗⃗⃗ ( 푟푒푓푙푒푥𝑖푣푒)

• If 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗ then 퐶퐷⃗⃗⃗⃗⃗ = 퐴퐵⃗⃗⃗⃗⃗ (푠푦푚푚푒푡푟𝑖푐) • If 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗ then 퐶퐷⃗⃗⃗⃗⃗ = 퐸퐹⃗⃗⃗⃗⃗ then 퐴퐵⃗⃗⃗⃗⃗ = 퐸퐹⃗⃗⃗⃗⃗ (푡푟푎푛푠𝑖푡𝑖푣푒) are colinear.

Theorem If 퐴퐵⃗⃗⃗⃗⃗ is a vector and there is a point 푃, then there is a unique point Q such that 푃푄⃗⃗⃗⃗⃗ = 퐴퐵⃗⃗⃗⃗⃗ .

Proof :

Suppose R is the midpoint of 퐵푃⃗⃗⃗⃗⃗ , 푄 = 푆푅(퐴) then 퐴퐵⃗⃗⃗⃗⃗ = 푃푄⃗⃗⃗⃗⃗ 표푟 푃푄⃗⃗⃗⃗⃗ = 퐴퐵⃗⃗⃗⃗⃗ .

To prove the uniqueness of point, let 퐴퐵⃗⃗⃗⃗⃗ = 푃푇⃗⃗⃗⃗⃗ .

Then 푆푅(퐴) = 푇, because R is thee midpoint of 퐵푃⃗⃗⃗⃗⃗ and the image of 퐴 by 푆푅, then 푇 = 푄 meaning that 푃푄⃗⃗⃗⃗⃗ is the only vector where 푃 is the initial point and 푄 is the terminal point which is equivalent to 퐴퐵⃗⃗⃗⃗⃗

Corollary 1:

If 푃1(푥1, 푦1), 푃2(푥2, 푦2), and 푃3(푥3, 푦3) are points with the specified

coordinates, then 푃(푥3 + 푥2 − 푥1 , 푦3 + 푦2 − 푦1) is the only point that satisfies

푃⃗⃗⃗3⃗⃗⃗푃⃗ = 푃⃗⃗⃗1⃗⃗푃⃗⃗⃗2

Corollary 2:

If 푃푛 = (푥푛, 푦푛), 푛 = 1,2,3,4 then 푃⃗⃗⃗1⃗⃗푃⃗⃗⃗2 = 푃⃗⃗⃗3⃗⃗푃⃗⃗⃗4 . If and only if 푥2 − 푥1 =

푥4 − 푥3, and 푦2 − 푦1 = 푦4 − 푦3.

The last concept that is related to the vector is a scalar multiplication by vector. Suppose that 퐴퐵⃗⃗⃗⃗⃗ is vector and 푘 is real number, so:

a. If 푘 > 0, then 푘퐴퐵⃗⃗⃗⃗⃗ is a vector 퐴푃⃗⃗⃗⃗⃗ which is defined as 퐴푃⃗⃗⃗⃗⃗ = 푘(퐴퐵⃗⃗⃗⃗⃗ ), 푃 ∈ 퐴퐵⃗⃗⃗⃗⃗ (ray 퐴퐵). b. If 푘 < 0, then 푘퐴퐵⃗⃗⃗⃗⃗ where 푃 is the opposite ray of 퐴퐵⃗⃗⃗⃗⃗ and 퐴푃⃗⃗⃗⃗⃗ = |푘|(⃗⃗⃗퐴퐵⃗⃗⃗⃗ )

B. Translation In this section, the concept of translation is introduced using the definition of vector.

Theorem

If 푠 and 푡 are two parallel lines, then 퐴 and 퐵 are the two points, so 퐴퐴⃗⃗⃗⃗⃗⃗⃗⃗′′ =

⃗⃗⃗⃗⃗⃗⃗⃗ ′′ ′′ 퐵퐵′′ with 퐴 = 푀푡푀푠(퐴) and 퐵 = 푀푡푀푠(퐵). Proof.

Choose a coordinate system with 푡 as the 푦-axis and a line which is perpendicular to 푡 as the 푥-axis, and s which its parallel to t.

Suppose 퐴 = (푎1, 푎2) and 퐵 = (푏1, 푏2). ⃗⃗⃗⃗⃗⃗⃗⃗ ′′ If 푁 𝑖푠 the midpoint of 퐴′′퐵, then we have to prove 푆푁(퐴) = 퐵 . If the equation of 푠 is 푥 = 푘(푘 ≠ 0), then ′′ 퐴 = 푀푠푀푡(퐴) = 푀푠(−푎1, 푎2) = (2푘 + 푎1, 푎2) ′′ 퐵 = 푀푠푀푡(퐵) = 푀푠(−푏1, 푏2) = (2푘 + 푏1, 푏2)

Because we know that 푁 is the midpoint of 퐴⃗⃗⃗⃗′′⃗⃗⃗퐵⃗ , then (2푘 + 푎 ) + 푏 푎 + 푏 푁 = [ 1 1 , 2 2] , whereas 2 2 2푘 + 푎 + 푏 푎 + 푏 푆 = [2 [ 1 1] − 푎 , 2 [ 2 2] − 푎 ] = (2푘 + 푏 , 푏 ) 푁 2 1 2 2 1 2 ′′ ′′ ⃗⃗⃗⃗⃗⃗⃗⃗ Evidently 푆푁(퐴) = 퐵 . Therefore 퐴⃗⃗⃗⃗퐴⃗⃗⃗⃗ = 퐵퐵′′

Definition

A mapping 퐺 is a translation, if there is a vector 퐴퐵⃗⃗⃗⃗⃗ such that for every point

′ ′ ⃗⃗⃗⃗⃗⃗⃗ ′ ⃗⃗⃗⃗⃗ 푃 in a plane 푉 has image 푃 푤here 퐺(푃) = 푃 and 푃푃 = 퐴퐵.

In the figure above, it is clear that every vector determines a translation.

If퐴퐵⃗⃗⃗⃗⃗ is a vector then 퐺퐴퐵 is a symbol to address a translation in the length of 퐴퐵.

Theorem

If 퐴퐵 = 퐶퐷 then 퐺퐴퐵 = 퐺퐶퐷

Proof:

If 푃 is an arbitrary point, it must be proved that 퐺퐴퐵(푃) = 퐺퐶퐷(푃). Suppose

퐺퐴퐵(푃) = 푃1 and 퐺퐶퐷(푃) = 푃2, so 푃푃1 = 퐴퐵⃗⃗⃗⃗⃗ and 푃푃2 = 퐶퐷⃗⃗⃗⃗⃗ . Since 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗ , 푃⃗⃗⃗⃗푃⃗⃗⃗1 =

푃⃗⃗⃗⃗푃⃗⃗⃗2 meaning that 푃1 = 푃2 and 퐺퐴퐵 = 퐺퐶퐷

Theorem

⃗⃗⃗⃗⃗ Suppose 푡 and 푠 as the two lines is parallel and 퐶퐷 is the vector that

perpendicular to 푠 and 푡, with 퐶 ∈ 푠 and 퐷 ∈ 푠. If 퐴퐵⃗⃗⃗⃗⃗ = 2퐶퐷⃗⃗⃗⃗⃗ so 퐺퐴퐵 = 푀푠푀푡

Proof:

Suppose 푃 is an arbitrary point, if 푃’ = 퐺퐴퐵 (푃) and 푃’’ = 푀푠푀푡 (푃), so it is necessary to prove that 푃’ = 푃’’

As the property of translation suggests, if 퐺퐴퐵(푃) = 푃’ then ⃗푃푃⃗⃗⃗⃗⃗ ’ = 퐴퐵⃗⃗⃗⃗⃗ . Since, 퐴퐵⃗⃗⃗⃗⃗ = 2퐶퐷⃗⃗⃗⃗⃗ , ⃗푃푃⃗⃗⃗⃗⃗ ’ = 2퐶퐷⃗⃗⃗⃗⃗ .

Related to 퐶’’ = 푀푠푀푡(퐶), 퐶 ∈ 푡 so 퐶’’ = 푀푡(퐶). It means that 퐷 is the midpoint of 퐶퐶⃗⃗⃗⃗⃗⃗⃗’’ implying that 퐶퐶⃗⃗⃗⃗⃗⃗⃗’’ = 2 퐶퐷⃗⃗⃗⃗⃗ . Since 퐶퐶⃗⃗⃗⃗⃗⃗⃗’’ = 푃푃⃗⃗⃗⃗⃗⃗⃗’’ , 푃푃⃗⃗⃗⃗⃗⃗⃗’’ = 2 퐶퐷⃗⃗⃗⃗⃗ = ⃗푃푃⃗⃗⃗⃗⃗ ’, and it means that 푃’ = 푃’’, so 퐺퐴퐵 = 푀푠푀푡.

Notes

1) Each translation 퐺퐴퐵 can be written as the composition between two reflections in two lines which are perpendicular to ̅퐴퐵̅̅̅ and it is ½ 퐴퐵 in length. 2) If 퐴퐵 is a line and 퐶 is the midpoint of ̅퐴퐵̅̅̅ whereas 푡, 푠 and 푛 are the perpendicular

lines to 퐴퐵 in 퐴, 퐶 and 퐵 respectively, then 퐺푎푏 = 푀푠푀푡 = 푀푛푀푠

3) Since any translation can be written as a composition of two reflections, whereas

a reflection is a transformation is isometry then a translation is an isometry transformation.

Translation is a direct isometry

Theorem

If 퐺 is a translation, then (퐺 )−1 = 퐺 퐴퐵 퐴퐵 퐵퐴

Proof:

According to latest figure, 퐺퐴퐵 = 푀푠푀푡 = 푀푛푀푠

While 퐺퐵퐴 = 푀푡푀푠 = 푀푠푀푛 −1 −1 −1 −1 (퐺퐴퐵) = (푀푠푀푡) = 푀푡 . 푀푠 = 푀푡푀푠 = 퐺퐵퐴

−1 So, (퐺퐴퐵) = 퐺퐵퐴.

C. The Closeness of translation In the previous section, it is explained that a translation can be expressed in the form of a composition of two reflections. This section will describe that the composition of the two translations is a translation as well. The assertion refer to the following theorem:

Theorem

If 퐺퐴퐵 is a translation, C and D are points such that 퐴퐵⃗⃗⃗⃗⃗ = 퐶퐷⃗⃗⃗⃗⃗ , then 퐺퐴퐵 = 푆푝푆퐶.

Proof : Suppose that 푔 = 퐶퐷⃗⃗⃗⃗⃗ , 푘 ⊥ 푔 𝑖푛 퐶, 푚 ⊥ 푔 𝑖푛 퐷

퐴퐵 is a vector from 푘 to 푚, therefore 퐴퐵⃗⃗⃗⃗⃗ = 2퐶퐷⃗⃗⃗⃗⃗ , so 퐺퐴퐵 = 푆퐷푆퐶. While 푆퐷 =

푀푚푀𝑔 and 푆푐 = 푀𝑔푀푘.

Then 푆퐷푆퐶 = (푀푚푀𝑔)(푀𝑔푀푘) = 푀푚(푀𝑔푀𝑔)푀푘 = 푀푚푀푘. 푆표 퐺퐴퐵 = 푆퐷푆퐶.

Example 1 Given 퐴 = (3, −1), 퐵 = (1,7) and 퐶 = (4,2). Find the coordinate of a point

퐷 such that 퐺퐴퐵 = 푆퐷푆퐶.

Solution: Suppose that 퐸 is a point such that 퐶퐸 = 퐴퐵, then 퐸[4 + (1 − 3),2 + (7 − (−1))] or 퐸 = (2,10). If 퐷 is the midpoint 퐶퐸̅̅̅̅, then 퐷 = (3,6), implying that 퐶퐸⃗⃗⃗⃗⃗ = 2퐶퐷⃗⃗⃗⃗⃗ .

Thus, 퐴퐵⃗⃗⃗⃗⃗ = 2퐶퐷⃗⃗⃗⃗⃗ , which obtains 퐺퐴퐵 = 푆퐷푆퐶 where 퐷 (3,6).

Theorem

The composition of a translation and a halfturn is a halfturn.

Proof:

Suppose that 퐺퐴퐵 is a translation and 퐶 is any point and suppose that 퐸 is also a point so 퐶퐸⃗⃗⃗⃗⃗ = 퐴퐵⃗⃗⃗⃗⃗ . If 퐷 mid point 퐶퐸̅̅̅̅̅ then 퐶퐸⃗⃗⃗⃗⃗ = 2퐶퐷⃗⃗⃗⃗⃗ .

According to previous theorem 퐺퐴퐵 = 푆퐷푆퐶, then 퐺퐴퐵푆퐶 = 푆퐷푆퐶푆퐶 = 푆퐷퐼 =

푆퐷. So 퐺퐴퐵푆퐶 = 푆퐷.

As a result of the theorem above is:

If 푆퐴, 푆퐵 and 푆퐶 is half round, then 푆퐴푆퐵푆퐶 = 푆퐷, with 퐷 is points that satisfy 퐴퐷⃗⃗⃗⃗⃗ = 퐵퐶⃗⃗⃗⃗⃗ .

To declare the composition of two translations is a translation in Cartesian coordinates, consider the following theorem.

Theorem

If 퐺푂퐴 is a translation where the coordinates of point 0 and point 퐴 are respectively

(0,0) and (푎, 푏). 푇 is a transformation that maps each point 푃 (푥, 푦) to 푇 (푃) =

(푥 + 푎, 푦 + 푏) then 퐺 = 푇. 푂퐴

Proof:

For 푃(푥, 푦), 푇(푃) = (푥 + 푎, 푦 + 푏). Suppose that 푃’ = 퐺푂퐴(푃) then 푃푃’ = 푂퐴 implying that 푃’ = (푥 + 푎 − 0, 푦 + 푏 − 0) = (푥 + 푎, 푦 + 푏). Thus 푇(푃) =

퐺푂퐴(푃) for each 푃 ∈ 푉. In other words 퐺푂퐴 = 푇. Example 2

Suppose that 퐺퐴퐵 is a translation which takes point 퐴(2,3) to point 퐵(4,1) and 퐺퐶퐷 is a translation which takes point 퐶(−3,4) to point 퐵(0,3). If 푃(푥, 푦). Determine the coodinate of 퐺퐶퐷퐺퐴퐵(푃).

Solution: ⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗ Suppose that 0’ = 퐺퐴퐵(0) and 0’’ = 퐺퐶퐷(푂) then 푂푂′ = 퐴퐵⃗⃗⃗⃗⃗ and 푂푂′′ = 퐶퐷⃗⃗⃗⃗⃗ . Then 0’ = (0 + 4 − 2,0 + 1 − 3) = (2, −2) and 0’’ = (0 + 0 + 3,0 + 3 − 4) = (3, −1).

So 퐺퐴퐵(푃) = (푥 + 2, 푦 − 2) and 퐺퐶퐷(푃) = (푥 + 3, 푦 − 1).

Thus 퐺퐶퐷퐺퐴퐵(푃) = 퐺퐶퐷[(푥 ± 2, 푦 − 2)] = (푥 + 2 + 3, 푦 − 2 − 1) = (푥 + 5, 푦 − 3) Exercises 1. Suppose there are 4 points that are wirtten as 퐴, 퐵, 퐶 and 퐷 of which each pair of three points is not collinear. Skecth the following: a. Point E such that 퐶퐸⃗⃗⃗⃗⃗ = 퐴퐵⃗⃗⃗⃗⃗ b. Point F such that 퐷퐹⃗⃗⃗⃗⃗ = 퐵퐴⃗⃗⃗⃗⃗

c. 푆퐴(퐴퐵)

2. Given points 퐴(0,0), 퐵(5,3), and 퐶(−2,4). Find the coordinate : a. R so that such that 퐴푅⃗⃗⃗⃗⃗ = 퐵퐶⃗⃗⃗⃗⃗ b. S so that such that 퐶푆⃗⃗⃗⃗ = 퐴퐵⃗⃗⃗⃗⃗ c. T so that such that 푇퐵⃗⃗⃗⃗⃗ = 퐴퐶⃗⃗⃗⃗⃗

3. If 퐴(1,3), 퐵(2,7), dan 퐶(−1,4) are the vertices of parallelogram 퐴퐵퐶퐷. Determine the coordinate of the point 퐷.

4. Suppose there are two lines 푔 and ℎ whcih are parallel, point 푃 ∈ 푔, and point 푄 neither on 푔 nor on ℎ.

a. Sketch 푃’ = 푀ℎ푀𝑔(푃) and 푄’ = 푀𝑔푀ℎ(푄)

b. Prove that 푃푃⃗⃗⃗⃗⃗⃗ ′ = 푄푄⃗⃗⃗⃗⃗⃗⃗ ′

5. Given a line 푔 and circles 퐿1 and 퐿2. Line 푔 are not cuting the circles 퐿1 and 퐿2. Use a transformation to draw a square having vertices located on 푔, a vertex that is located

on 퐿1and the other vertex points located on 퐿2

6. If 푃0 = (0,0), 푃1 = (푥1, 푦1), 푃2 = (푥2, 푦2), and 푃3 = (푥3, 푦3) while 푘 > 0.

a. Find the coordinate of 푃 such that 푃⃗⃗⃗0⃗⃗⃗푃⃗⃗1 = 푘푃⃗⃗⃗0⃗⃗⃗푃⃗⃗1

b. Find the coordinate of 푃 such that 푃⃗⃗⃗1⃗⃗⃗푃⃗ = 푘푃⃗⃗⃗1⃗⃗푃⃗⃗⃗2

c. Is the statement “If 푃⃗⃗⃗3⃗⃗⃗푃⃗ = 푘푃⃗⃗⃗1⃗⃗푃⃗⃗⃗2 , then 푃 = {푥3 + 푘(푥2 − 푥1), 푦3 + 푘(푦2 −

푦1)}.” valid for 푘 < 0?

7. Given 퐴, 퐵, and 퐶 which are not collinear. Sketch:

a. 퐺퐴퐵(퐴) and 퐺퐴퐵(퐵).

b. 퐺퐴퐵(퐶).

c. The lines 푔 and ℎ where 퐴 ∈ 푔 and 퐺퐴퐵 = 푀ℎ푀𝑔.

8. Suppose there are two points that notated as 퐴 and 퐵 and line 푔 such that 푔 ⊥ 퐴퐵⃗⃗⃗⃗⃗ . Sketch:

a. Line ℎ such that 푀ℎ푀𝑔 = 퐺퐴퐵.

b. Line 푘 such that 푀𝑔푀푘 = 퐺퐴퐵.

c. Line 푚 such that 푚’ = 퐺퐴퐵(푚).

d. Point 퐶 such that the 퐺퐵퐴 (퐶) = 퐵

9. Suppose lines 푔 and ℎ are parallel and there is point 퐴 which is not on the line.

a. Draw the point 퐵 such that 푀ℎ푀𝑔 = 퐺퐴퐵

b. Draw the point 퐶 such that 푀𝑔푀ℎ = 퐺2퐴퐶

10. Given 퐴 (2,3) and 퐵 (−4,7), determine the equation of a line 푔 and ℎ such that

푀ℎ푀𝑔 = 퐺퐴퐵

11. Given three points 퐴 (−1,3), 퐵 (5, −1) and 퐶 (2,4)

a. Find the coordinate 퐶 ′ = 퐺퐴퐵 (퐶)

b. Find the equation of lines 푔 and ℎ so 퐶 ∈ 푔 and so 푀ℎ푀𝑔 = 퐺퐴퐵

12. The edges of a river are depicted with two parallel lines that is written as 푡 and 푠 (see figure). Above the river, there will be built a bridge and according to a good construction, the bridge must be made perpendicular to the direction of the river. Where is the bridge should be constructed usch that the distance fromthe town 퐷 to the town 퐸 will be as short as possible.

CHAPTER VII ROTATION A. Directed Angle Angle has been introduced previously as the combination between two rays that have identical initial point. For example, angle ABC notated by ∠ ABC is formed by BA and BG rays. We can see angle ABC in the following figure:

Both figures (a and b) certainly illustrate angle ABC. But for the next discussion both figures will be distinguished. It is distinguished by using initial ray and terminal ray of an angle. It’s used to determine what type of positive angle or negative angle from an angle. Such angle is called directed angle

Definition

Directed angle is an angle of which one of the ray is the initial ray and the other is the terminal ray

To symbolize an angle, for example ∠ ABC is a directed angle where 퐵퐴⃗⃗⃗⃗⃗ rays is the initial ray and 퐵퐶⃗⃗⃗⃗⃗ is the terminal ray notated as ∠ ABC For another purpose,

When < 퐴퐵퐶 is an angle then < 퐴퐵퐶 = < 퐶퐵퐴, so 푚 < 퐴퐵퐶 = −푚 < 퐶퐵퐴. But for a directed angle < 퐴퐵퐶 it applies for the 푚 < 퐴퐵퐶 = −푚 퐶퐵퐴 since the orientation of the 퐵퐴퐶 is always contrary to the orientation of the 퐵퐶퐴. If there are two intersecting lines not perpendicular to each other, then its angle is the acute angle.

In the figure, the magnitude of the angle between line g and k is 80 °, and the magnitude of the angle between h and k is -60 °. The angle between two lines can be described as follows: Suppose that g and h intersect at point O, point P is on line g, points Q and R are located on line h as suggested in the following figure:

If ∠POR is an acute angle, then the magnitude of line g to line h is equal to ∠POR, whereas if the angle ∠POR is an obtuse angle, then the angle from line g to line h equals the magnitude of ∠POQ. Suppose m∠POR = 140 °, then the magnitude from line g to line h is m∠POQ = 40 °, while the magnitude of line h to line g is m∠QOP = -40 °.

B. Rotation It will be discussed in the present section, the composition of the two reflection in the lines that are not parallel and in intersecting lines but are not perpendicular to each other. The composition of the two reflections will produce an isometry either in the form of a rotation and a translation. The composition is a basic theorem of rotation.

Theorem B If s and t intersect at point A but are not perpendicular, the points P and Q are the points that are different from A, then m (< PAP”) = m (< QAQ”) with P” =MtMs(P) and Q” =MtMs(Q)

Proof:

Case 1 : P∈ s and Q∈ s

A”= MtMs(A) means that A”=A. P”=MtMs(P) and Q”=MtMs(Q). Sincethe points A, P and Q are located on are collinear, A”, P”, and Q” are collinear. Thus, lines PP” and QQ” intersect at point A. Thus m (< PAP”) = m (< QAQ”).

Case 2 : P∉ s and Q∈ s

m ( < PAP”) = m ( < PAQ ) + m ( < QAP”), then m (< QAQ”) = m (< QAP”) + m (< P”AQ”), since m (< PAQ)= m (< P”AP”) so m (< PAP”) = m (< QAQ”).

Case 3 : P∉ s and Q∉ s

In the case 3, we can prove that if P”= MtMs(P) and Q”=MtMs(Q), so m (< QAQ” = m (< PAP”).

Therefore, by transformation of MtMs, every point rotates by the same directed angle, rotating the same point. Such process is called a rotation.

Defenition If A is a point and 훼 is angle where -180°<훼<180°, a rotation with center A and angle 훼 denoted by 푅퐴,훼 is a function from V to V defined by

1. If P = A, so 푅퐴,훼(P) = P.

2. If P ≠ A, so 푅퐴,훼(P) = P’, then m (< PAP”) = 훼and AP’=AP.

Based on the definition, the rotation 푅퐴,훼, just has an invarioant point, i.e. A (the center of the rotation). The image of point the P by rotation 푅퐴,훼, is a point of a circle where A is the center and AP is the radius. Since magnitude α of the angle rotation is between -1800 and 1800, so α>0, if the direction of the angle is opposite to the clockwise direction, and α<0 if the direction of the angle is the same as the of the clockwise direction.

RA, α(P)

A P

RA, -α(P)

Theorem If s and t are two lines that intersect on A, s and t aren’t perpendicular, and if the magnitude of the angle from line s to t is α/ , then R =M M 2 A,α t s

Proof:

K t

½ α K s A

Suppose that a point K ≠ A located on s. If K’ = MtMs(K) so m(

Since

Theorem

1. Rotation with A as the center and the magnitude α (RA,α) is a transformation. 2. Every rotation is direct isometry 3. The composition of two reflections is either rotation or translation

Example

If RA,α is a rotation that maps point P to P’, determine two pairs of lines which can be used as some reflection axes such that the composition of these reflections is a rotation.

Solution

½ α

½ α u A P

v 1. Suppose that s = 퐴푃̅̅̅̅, t is the bisecting line of

angle from s to t is ½ α, so RA,α = MtMs.

2. Suppose u = 퐴푃̅̅̅̅̅′, and v is a line passing through A, then the angle magnitude from u to

v is ½ α, so RA,α = MvMu.

C. Rotation Composition We have proved that the rotation is a transformation. Because the composition of two transformations is transformation, what about the composition of two rotations? Whether it is a rotation or another transformation? For the composition of two transformations, the following matters are discussed: a. If the centers of the rotations are the same b. If the centers of rotation are different If the centers of the rotations are the same, the composition of two rotations is a rotation with the same center and the magnitude of the angle is the sum of the angles (provided that if the sum is greater than 180° it must be substracted by 360°while if the sum is less than -180° it must be added by 360°). Meanwhile, the composition of the rotations having different centers, can be in the form of a rotation of which the center is different to the centers of th composed rotations . In addition, the angle magnitude is the sum of the angles of the rotations following the previous case order. If the sum of the angles of is zero, then the composition of two rotations forms a translation. In general it can be conluded as follows:

If RA,훼1 andRB,훼2 is two rotation, then RB,훼2RA,훼1 = Rc,훼 with the conditions of 훼 as follows: 1. If 0° < |훼1 + 훼2| < 180° 푡ℎ푒푛 훼 = 훼1 + 훼2. 2. If 훼1 + 훼2 > 180° 푡ℎ푒푛 훼 = (훼1 + 훼2) − 360°. 3. If 훼1 + 훼2 < −180° 푡ℎ푒푛 훼 = (훼1 + 훼2) + 360°.

Example

RA,120 . RA,30= RA,150

RB,160 . RB,40= RB,-160 .

RC,-150 .RC,-50 .= RC,160 .

While if훼1 + 훼2 = 0°, 푡ℎ푒푛 RB,훼2RA,훼1is a rotation .

Theorem

The composition of two rotations is a rotation or translation.

Proof :

Assume there are rotations 푅퐴,훼and 푅퐵,훽. Draw a line s = 퐴⃡⃗⃗⃗퐵⃗ , if m(∠XAY) = m(∠XAZ) 1 = 훼, then 푅 = 푀 푀 and 푅 = 푀 푀 . 2 퐴,훼 푠 푡 퐵,훽 푢 푠 So 푅퐵,훽푅퐴,훼 = 푀푢푀푠푀푠푀푡= 푀푢푀푡.

If u is parallel to t then 푅퐵,훽푅퐴,훼 is a translation, and if u and t intersect at C, then

푀푢푀푡 is a rotation centered at C. Assume 푅퐶,0 = 푅퐵,훽푅퐴,훼then in 훼, 훽, dan 휃, there is following relationship: 1 1 m(∠ABC) = 훼, m(∠BAC) = 훽. 2 2 1 1 1 1 Therefore m(∠PCB) = 훼 + 훽. It means that the angle from t to u is 훼 + 훽, so that 2 2 2 2 휃 = 훼 + 훽. If 훼 + 훽 > 180°, then 휃= (훼 + 훽) − 360°.

EXERCISES 1. Given points A and P are different. Construct the following A. RA,90 (P) B. RA,150 (P) C. RA,45 (P) D. Q such that RA,30 (Q) = P 2. Given m (

a. RA,30, RA,90 b. RA,-60, RA,120 c. RA,135, RA,90 d. RA,-120 RA,-150 4. Simplify the following transformation compositions: A. RA,30 RA,60

B. RA,120 RA,-90 C. RA,135 RA,-90 D. RA,-60 RA,45 E. RA,-120 RA,-150 F. RA,-120 RA,90 5. Suppose there are two lines s and t which intersect at point A and two points P and Q not on the lines. ’ A. Construct point P = MtMs(P) ’ B. Construct point Q = MsMt(Q) ” C. Construct point P = MtMs(P) D. if m (

a. 푅0,90(퐴) b. 푅0,45(퐴) c. 푅0,120(퐴)

d. 푅0,−135(퐴) 8. Writen the equation of lines s and t such that 푀푠푀푡 equals the following rotation, if A=(1,3) and 0 is the center point.

a. 푅푂,90 b. 푅푂,−180 c. 푅푂,120 d. 푅퐴,90 e. 푅퐴,−90 9. If A is the center point of an orthogonal coordinate system and 푠 = {(푥, 푦)|푦 = 2푥 − ’ 3} find the equation of s = 푅퐴,90(푆) 10. If I is a circle with radius 2 and the center in A=(√2, √2) and if B = (0,0), find the

equation of I’=푅퐴,90(퐼)

REFERENCES

Remsing, Claudia. 2006. Transformation Geometry. Rhodes University