Diktat Geometri Transformasi (English)

i
TRANSFORMATION GEOMETRY MODULE
FAJAR ARWADI
MATHEMATICS DEPARTMENT
MATHEMATICS AND NATURAL SCIENCE FACULTY
UNIVERSITAS NEGERI MAKASSAR
2018

ii
QUOTES
The Prophet Muhammad (‫ﷺ‬ – peace be upon him) said: “Acquire
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

iii
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.

iv
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

1
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:
In general, a function is notated with the letter 𝑓. If 𝑓 is a function from V to V that associates
every x ∈ V to y ∈ V, it can be written as 𝑦 = 𝑓( 𝑥) where 𝑥 is called the pre-image of 𝑦 by 𝑓
and 𝑦 is called the image of 𝑥 by 𝑓. The origin and the range of the function are 𝑉.
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 𝑓( 𝑥) = 𝑦. 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 𝑓 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.
Definition:
A function of V to V is a mapping that associates every element of V to exactly one element
of V
.

2
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.
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.
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

3
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 𝐴𝑃̅̅̅̅.
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 ∈ 𝐴𝑅̅̅̅̅
, 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.

4
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 ∈ 𝐴𝑌̅̅̅̅ 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′ ∈ 𝐴𝑃̅̅̅̅ and Q′
∈ 𝐴𝑄̅̅̅̅, then 𝐴𝑃̅̅̅̅ and 𝐴𝑄̅̅̅̅ has two intersection points, namely point A
and point P’ or Q’. It means 𝐴𝑃̅̅̅̅ and 𝐴𝑄̅̅̅̅ coincide Since Q ∈ 𝐴𝑄̅̅̅̅ and 𝐴𝑃̅̅̅̅ = 𝐴𝑄̅̅̅̅ then Q ∈ 𝐴𝑃̅̅̅̅,
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.

5
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 𝐴(x, y), is 𝐴 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)
we get 𝑥′
=
𝑥−1
2
𝑎𝑛𝑑 𝑦′
=
2𝑦+𝑥−1
2
.
So B = (
x−1
2
,
2y±1
2
), hence T((
x−1
2
,
2y±1
2
) = (x, y).
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 𝑥1 = 𝑥2 and 𝑦1 = 𝑦2 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

6
●A
Excercises
1) Let point K and line segment of 𝐴𝐵̅̅̅̅ where K ∉ 𝐴𝐵̅̅̅̅ and a line 𝑔 so that 𝑔 // 𝐴𝐵̅̅̅̅ and the
distance between K and 𝐴𝐵̅̅̅̅ is two times the distance between K and 𝑔. There is a mapping
T with domain 𝐴𝐵̅̅̅̅ and the range 𝑔 so that if P ∈ 𝐴𝐵̅̅̅̅ then T(P) = P’ = 𝐾𝑃̅̅̅̅ ∩ g.
a. What is the range of P’ if P moves through 𝐴𝐵̅̅̅̅.
b. Prove that T is injective.
c. If E and F are two points on 𝐴𝐵̅̅̅̅, what can be interpreted about the distance of E’F’, if
E’ = T(E) and F’ = T(F).
2) Let O(0,0), C1 = {(x, y)│x2
+ y2
= 1}, C2 = {(x, y)│ x2
+ y2
= 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) 𝐿𝑒𝑡 𝐹 ∶ 𝑉 −> 𝑉 , 𝑖𝑓 𝑃 (𝑥, 𝑦) 𝑡ℎ𝑒𝑛 𝑓(𝑃) = (│𝑥│, │𝑦│)
a. Determine 𝑓(𝐴) if 𝐴(−,6)
b. Determine all the pre-images of the point 𝐵(4,2)
c. What is the shape of the range?
4) Let function 𝑓 ∶ axis 𝑥 −> 𝑣 defined as: “if 𝑃(𝑥, 0) then 𝑓(𝑃) = (𝑥, 𝑥2
)
a. Find the image of 𝐴(5,0) by 𝑓
b. Is 𝐵(−13,169) ∈ image of 𝑓
c. Is f surjective?
5) Let a line 𝑠 and point of 𝐴, 𝐵, 𝐶 as shown below
𝑇: 𝑉 → 𝑉 is defined as follows:
i) if 𝑃 ∈ 𝑆 then ( 𝑃) = 𝑃
ii) if 𝑃 ∉ 𝑆 then 𝑇(𝑃) = 𝑝’ such that is axis of 𝑃𝑃′̅̅̅̅̅
●C
●B

7
a. Draw 𝐴’ = 𝑇(𝐴), 𝐵’ = 𝑇(𝐵)
b. Draw the pre-image of point 𝐶
c. is 𝑇 a tranformation?
d. Prove that 𝐴’𝐵’ = 𝐴𝐵
6) Let two lines 𝑔 and h are parallel in the Euclidean plane 𝑉, and a point of 𝐴 which is in the
middle between 𝑔 and ℎ. T is the image of the domain of g defined as : 𝑖𝑓 𝑃 ∈ 𝑔 𝑡ℎ𝑒𝑛 𝑃’ =
𝑇(𝑃) = 𝑃𝐴̅̅̅̅ ∩ ℎ
a. What is the range of T
b. If 𝐵 ∈ 𝑔, 𝐶 ∈ 𝑔, and 𝐵 ≠ 𝐶, 𝐵’𝐶 = 𝐵𝐶 𝑤𝑖𝑡ℎ 𝐵’ = 𝑇(𝐵), 𝐶’ = 𝑇(𝐶)
c. is 𝑇 injective?
7) Let three different points 𝐴, 𝐸, 𝐷 not collinear and a relation 𝑇 defined as:
𝑇(𝐴) = 𝐴, 𝑇(𝑃) = 𝑃’, such that P is the middle point of 𝐴𝑃̅̅̅̅.
a. Draw 𝐸’ = 𝑇(𝐸)
b. Draw 𝑄 so that 𝑇(𝑄) = 𝐷
c. is 𝑇 a transformation?

8
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:
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 𝐴𝐴′̅̅̅̅̅ (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 𝑀𝑠(𝐴) = 𝐴’ = 𝐴 and Ms(B) = B’ such that s is the axis 𝐵𝐵′̅̅̅̅̅.
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 𝐴𝐴′̅̅̅̅̅ ⊥ s and 𝐵𝐵′̅̅̅̅̅ ⊥ s , so 𝐵𝐴′̅̅̅̅̅ ⊥ 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’.
Definition:
Reflection in line 𝑠 is a function 𝑀𝑠 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 𝑃𝑃′̅̅̅̅̅.

9
Then, M is injective.
Since MS is a function that is both surjective and injective, Ms is 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 𝑃𝑃′̅̅̅̅̅ ⊥ s , then:
𝑦′−𝑦
𝑥′−𝑥
=
𝑏
𝑎
.................... (i)
If Q is the midpoint of 𝑃𝑃′̅̅̅̅̅ , then Q(
𝑥+𝑥′
2
,
𝑦+𝑦′
2
) lies on the line s.
So a(
𝑥+𝑥′
2
) + b(
𝑦+𝑦′
2
) + c = 0.................(ii)
From the equation (i) and (ii), it is obtained that:
x’ = x -
2𝑎(𝑎𝑥+𝑏𝑦+𝑐)
𝑎2+ 𝑏2
y’ = y -
𝑎2+ 𝑏2
so, if s: ax + by + c = 0, and P(x,y), then Ms(P) = P’(x’,y’) where
x’ = x -
𝑎2+ 𝑏2
y’ = y -
𝑎2+ 𝑏2
Theorem:
Every reflection in a line is a transformation

10
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 𝐴𝐵̅̅̅̅. If T is the midpoint of 𝐴𝐵̅̅̅̅, then s passes through the point
T and perpendicular to 𝐴𝐵̅̅̅̅. m𝐴𝐵⃡⃗⃗⃗⃗ =
2
3
=> mS = -
3
2
, and T (-
1
2
, 2).
So, the line s : y – 2 = -
3
2
(x +
1
2
) or
s : 3x + 2y - 2
1
2
= 0 or s : 6x + 4y – 5 = 0
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’)
x’ = x -
2𝑎( 𝑎𝑥+𝑏𝑦+𝑐)
𝑎2+ 𝑏2
x’ = 2 -
2.2[2.2+(−3).(−5)+5]
22+ (−3)2
= 2 -
96
13
=
−70
13
y’ = y -
2𝑏( 𝑎𝑥+𝑏𝑦+𝑐)
𝑎2+ 𝑏2
y’ = -5 -
2.(−3)[2.2+(−3).(−5)+5]
22+ (−3)2
= -5 -
−144
13
=
79
13
Thus, Ms(A) = A’ (
−70
13
,
79
13
)
c. Ms(O) = O’(x’,y’)
x’ = x -
2𝑎( 𝑎𝑥+𝑏𝑦+𝑐)
𝑎2+ 𝑏2

11
= 0 -
2.2[2.0+(−3).0+5]
22+ (−3)2
= 0 -
20
13
=
−20
13
y’ = y -
2𝑏( 𝑎𝑥+𝑏𝑦+𝑐)
𝑎2+ 𝑏2
= 0 -
2.(−3)[2.0+(−3).0+5]
22+ (−3)2
= 0 -
(−30)
13
=
30
13
Thus, Ms(O) = O’ (
−20
13
,
30
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 :

12
l
B t
A
a. Draw Mt(∆ ABC)
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
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.
● D
C
A●
● B

13
k
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).
A●
l

14
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 𝑇 is an isometry if for every pair of points 𝑃, 𝑄 satisfies 𝑃′
𝑄′
=
𝑃𝑄, where 𝑃′
= 𝑇(𝑃) and 𝑄′
= 𝑇(𝑄).
If 𝐴′
= 𝑀𝑠(𝐴), 𝐵’ = 𝑀𝑠(𝐵), then 𝐴’𝐵’ = 𝐴𝐵. (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 aline into a line
Suppose g is a line and 𝑇 is an isometry, it will be proved that 𝑔′ = 𝑇(𝑔) is a line.
Put the points 𝐴 and 𝐵 on the line 𝑔 (𝐴 ∈ 𝑔 and 𝐵 ∈ 𝑔).
Suppose 𝑇(𝐴) = 𝐴′ and 𝑇(𝐵) = 𝐵′, create a line h through point 𝐴′and 𝐵′. It will be
proved that the line ℎ = 𝑔′
Take an arbitrary point 𝑃 on 𝑔 such that it forms 𝐴𝑃𝐵, and let 𝑃′ = 𝑇(𝑃). On the line 𝑔,
𝐴𝑃 + 𝑃𝐵 = 𝐴𝐵.
Because 𝑇 is an isometry then 𝐴′𝐵′ = 𝐴𝐵, 𝐴′𝑃′ = 𝐴𝑃, 𝑃′𝐵′ = 𝑃𝐵.
Suppose 𝑃’ lies on outside of ℎ, then at ∆𝐴′𝑃′𝐵′ must be satisfied that 𝐴′𝑃′ + 𝑃′𝐵′ > 𝐴′𝐵′
since 𝐴′𝐵 ′ = 𝐴𝐵, 𝐴′𝑃′ = 𝐴𝑃, 𝐴𝑃 + 𝑃𝐵 > 𝐴𝐵

15
It contradicts to the assumption that 𝐴𝑃 + 𝑃𝐵 = 𝐴𝐵.
It means that the assumption that 𝑃′ lies on outside of the h is not true. Suppose 𝑃′ lies
on ℎ or 𝐴′𝑃′𝐵′.
So 𝑔’ ⊂ ℎ
The reverse direction is proved in the same way by assuming that Q' is
any point on h.
Since 𝑇 is a transformation, it means that it satisfies the surjective properties, then there
is a 𝑄 such that 𝑇(𝑄) = 𝑄′. Suppose 𝑄 lies outside of 𝑔. Using the triangle inequality, it
can be proved that 𝑄 should be on g, so that 𝑄′
= 𝑇(𝑄) must be on 𝑔′
= 𝑇 (𝑔). Thus, it
means that ℎ′ ⊂ 𝑔′.
Since 𝑔′ ⊂ ℎ and ℎ ⊂ 𝑔′ then ℎ = 𝑔′
b. Preserving the size of the angle between two lines
Take the three points 𝐴, 𝐵, and 𝐶 which are not collinear.
𝐴′ = 𝑇(𝐴), 𝐵′ = 𝑇(𝐵) and 𝐶′ = 𝑇 (𝐶)
See the 𝛥𝐴𝐵𝐶. According to (a), since the AB and BC are straight lines so 𝐴′𝐵′ and
𝐵′𝐶′ are also straight lines.
Because 𝑇 is an isometry then 𝐴′𝐵′ = 𝐴𝐵, 𝐵′𝐶′ = 𝐵𝐶, and 𝐴′𝐶′ = 𝐴𝐶.
It means that 𝛥𝐴𝐵𝐶 ≅ 𝛥𝐴′𝐵′𝐶 ′(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.

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c. Preserving the parallels of two lines
Given Line 𝑔 // ℎ, 𝑔 ′ = 𝑇 (𝑔) and ℎ′ = 𝑇 (ℎ).
It will be proved that 𝑔 ′// ℎ′
Suppose the line 𝑔 'intersect ℎ′ at the point 𝑃’, then 𝑃′ ∈ 𝑔′ and 𝑃′ ∈ ℎ′. Since 𝑇 is a
transformation then for 𝑃′ ∈ ℎ′ there must exist 𝑃 so that 𝑇(𝑃) = 𝑃′ where 𝑃 ∈ 𝑔 and
𝑃 ∈ ℎ. It then implies 𝑔 and ℎ intersect at point 𝑃.
It contradicts to the assumption that 𝑔 // ℎ implying the assumption that 𝑔 'intersects ℎ′
is wrong, it should be 𝑔′// ℎ′. One consequence of that property is that if 𝑔⟘ℎ then
𝑔′⟘ℎ ′, where T is an isometry, 𝑔′ = 𝑇(𝑔), and ℎ′ = 𝑇(ℎ). If 𝛥𝐴𝐵𝐶 are reflected to the
line g, the map is 𝛥𝐴′𝐵′𝐶′, or 𝑀𝑔(𝛥𝐴𝐵𝐶) = 𝛥𝐴′𝐵′𝐶′
A reflection in the line 𝑔 maps the 𝛥𝐴𝐵𝐶 on 𝛥𝐴′𝐵′𝐶′. If the 𝛥𝐴𝐵𝐶 with the order of the
𝐴 − 𝐵 − 𝐶 is opposite to the clockwise direction, then the image which is 𝛥𝐴′𝐵′𝐶’ having
the order 𝐴′ − 𝐵′ − 𝐶′ is clockwise direction.

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Definition
i. A transformation 𝑇 maintains an orientation if for any three points which
are not collinear (𝑃, 𝑄, 𝑅) have the same orientation as the orientation of
the triple points (𝑃′, 𝑄′, 𝑅′)
ii. A transformation T reverses an orientation if the orientation of any three
points which are not collinear (𝑃, 𝑄, 𝑅) is not the same as the orientation
of (𝑃′
, 𝑄′
, 𝑅′) where 𝑃′ = 𝑇(𝑃), 𝑄′
= 𝑇(𝑄), and 𝑅′ = 𝑇(𝑅).
Meanwhile, the rotation about the center of rotation 𝑂 maps 𝛥𝑃𝑄𝑅 on 𝛥𝑃′𝑄′𝑅′. If on the
𝛥𝑃𝑄𝑅, the direction of 𝑃 − 𝑄 − 𝑅 is clockwise direction then its image 𝛥𝑃′𝑄′𝑅′, has also
clockwise direction 𝑃′ − 𝑄′ − 𝑅′.
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 (𝐴, 𝐵, 𝐶) is the pair three points which are not collinear. Then through 𝐴,
𝐵, and 𝐶, there is exactly one circle 𝐼. We can circumnavigate 𝐼 started from 𝐴, then in
𝐵, then in 𝐶 and ended up back at 𝐴.
If the circumferential direction is the same as the clockwise direction, it is said
that the pair of three points (𝐴, 𝐵, 𝐶) has opposite orientations to clockwise direction or
negative orientation.
It means that the reflection of the line g that maps 𝛥𝐴𝐵𝐶 into the 𝛥𝐴′𝐵′𝐶′,
(𝐴, 𝐵, 𝐶) has negative orientation, and (𝐴′, 𝐵 ′, 𝐶′) has positive orientation. While in the
rotation with the center of rotation 𝑂 which maps 𝛥𝑃𝑄𝑅 into the 𝛥𝑃′𝑄′𝑅′, (𝑃, 𝑄, 𝑅) has
a positive orientation and (𝑃′, 𝑄 ′, 𝑅′)’s orientation is also positive.
Therefore
Moreover, we can classify isometry into direct isometry and indirect isometry (opponent
isometry) by looking at the following definition:

18
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 :
𝑇 is a transformation defined by 𝑇 (𝑃) = (𝑥 − 7, 𝑦 + 4) for all points 𝑃 (𝑥, 𝑦) ∈ 𝑉.
Show whether 𝑇 is an isometry?
Solution:
Suppose that 𝑃’ = 𝑇(𝑃) = (𝑥 − 7, 𝑦 + 4) ∀ 𝑃(𝑥, 𝑦) ∈ 𝑉
Suppose also that Point 𝐴(𝑥1, 𝑦1) and (𝑥2, 𝑦2) with 𝐴 ≠ 𝐵.
It means that 𝐴’ = 𝑇(𝐴) = ( 𝑥1 − 7, 𝑦1 + 4) and
𝐵’ = 𝑇(𝐵) = (𝑥2 − 7, 𝑦2 + 4)
𝐴𝐵 = √( 𝑥2 − x1)2 + ( 𝑦2 − 𝑦1)2
𝐴′𝐵′ = √{( 𝑥2 − 7) − (x1 − 7)}2 + {( 𝑦2 + 4) − ( 𝑦1 + 4)}2
= √( 𝑥2 − x1)2 + ( 𝑦2 − 𝑦1)2
Since 𝐴’𝐵’ = 𝐴𝐵, so 𝑇 isometry
Example 2
Suppose 𝑇 is a transformation that is defined for all points 𝑃(𝑥, 𝑦) as 𝑇(𝑃) = (− 𝑦, 𝑥).
a. Is 𝑇 an isometry?
b. If 𝑇 isometry, is it direct isometry or opponent isometry? Answer:
Solution:
a. Given 𝑃’ = 𝑇(𝑃) = (𝑦, −𝑥) ∀ 𝑃(𝑥, 𝑦) ∈ 𝑉
Suppose that point 𝐴(𝑥1, 𝑦1) and (𝑥2, 𝑦2) with 𝐴 ≠ 𝐵.
𝐴𝐵 = √( 𝑥2 − x1)2 + ( 𝑦2 − 𝑦1)2
𝐴′𝐵′ = √( 𝑦2 − 𝑦1)2 + {(−𝑥2) − (−x1)}2
= √( 𝑦2 − 𝑦1)2 + (−𝑥2 + x1)2
= √( 𝑥2 − x1)2 + ( 𝑦2 − 𝑦1)2
evidently 𝐴’𝐵’ = 𝐴𝐵

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Because 𝐴’𝐵’ = 𝐴𝐵, so 𝑇 isometry.
b. To check whether 𝑇 is a direct isometry or opponent isometry, take any three points
which are not collinear, for example 𝑂(0.0), 𝐴(1,2) and 𝐵(4.1).
By using transformation 𝑇(𝑃) = (𝑦, −𝑥) ) ∀ 𝑃(𝑥, 𝑦) ∈ 𝑉 then it is obtained that 𝑂’ =
(0,0), 𝐴’ = (−2,1), and 𝐵’ = (−1,4).
Since the orientation (𝑂, 𝐵, 𝐶) is positive, and the orientation (𝑂′, 𝐴′, 𝐵′) is also
positive, then 𝑇 is a direct isometry.
Exercises
1. 𝑇 is a transformation defined by 𝑇(𝑃) = (3𝑥 + 5,2 − 4𝑦) for all points 𝑃(𝑥, 𝑦) ∈
𝑉.
Show that T is an isometry.
2. Suppose the points 𝐴(1, −1), 𝐵(4,0), 𝐶(−4,1) and 𝐷(−2, 𝑝). If an isometry 𝑇 with
𝑇(𝐴) = 𝐶 and 𝑇(𝐵) = 𝐷, find the value of 𝑝.
3. Prove that the transformation with the formula:
[
𝑥′
𝑦′
] = [
3/5 −4/5
−4/5 −3/5
] [
𝑥
𝑦] is an isometry.
4. Suppose 𝛥𝐴𝐵𝐶 by isometry 𝑇 is mapped into 𝛥𝐴′𝐵′𝐶′, prove that 𝛥𝐴𝐵𝐶 ≅
𝛥𝐴′𝐵′𝐶′.
5. Given a line 𝑔. 𝑇 a function defined for each point of 𝑃 in the plane 𝑉 as follows:
i. If 𝑃 ∈ 𝑔 then 𝑇(𝑃) = 𝑃
ii. If 𝑃 ∉ 𝑔 then 𝑇(𝑃) = 𝑃’ such that 𝑃′ is the midpoint of the line segment of
orthogonal from 𝑃 to 𝑔.
a. Is 𝑇 a transformation?
b. Is 𝑇 an isometry?
c. If there are two points 𝐴 and 𝐵 so that 𝐴′𝐵 ′ = 𝐴𝐵 where 𝐴′ = 𝑇 (𝐴),
𝐵 ′ = 𝑇 (𝐵), what can be interpreted about 𝐴 and 𝐵?
6. Suppose the 𝑆 line and the points 𝐴, 𝐴′, 𝐵 and 𝐶 with 𝐴′ = 𝑀𝑔(𝐴). By only using a
ruler without ascale, sketch point 𝐵’ = 𝑀𝑔(𝐵) and 𝐶’ = 𝑀𝑔(𝐶).

20
7. Given lines 𝑠, 𝑡, 𝑢; points 𝐴 and 𝐵 as in the following figure. 𝑇 is an isometry
where 𝐵 = 𝑇(𝐴) and 𝑢 = 𝑇(𝑠). If 𝑡⟘𝑠, painting 𝑡′ = 𝑇(𝑡)
8. Suppose a line 𝑔 and and a circle 𝐼. Prove that 𝑀𝑔(𝐼) = 𝐼’ with 𝐼’ is also a circle.
9. Suppose lines 𝑔, 𝑔’, ℎ, ℎ’ and 𝑘 where 𝑔’ = 𝑀𝑘(𝑔) and ℎ’ = 𝑀𝑘(ℎ). If 𝑔’//ℎ’ prove
that 𝑔//ℎ.
10. Suppose lines 𝑔, ℎ, and ℎ’ where ℎ’ = 𝑀𝑔(ℎ). Verifying the truth of the following
expressions:
a. If ℎ’//ℎ, then ℎ//𝑔
b. If ℎ’ = ℎ, then ℎ = 𝑔.
c. If ℎ’⋂ ℎ = {𝐴}, then 𝐴 ∈ 𝑔.
11. Suppose line g and two points 𝐴 and 𝐵 as shown in the following figure:
a. by using an appropriate isometry, determine a point 𝑃 ∈ 𝑔 so that 𝐴𝑃 + 𝑃𝐵 as
short as possible.
b. If 𝑄 ∈ 𝑔 is different to the point 𝑃, prove that 𝐴𝑄 + 𝑄𝐵 > 𝐴𝑃 + 𝑃𝐵.
12. Suppose a circle 𝐼 = {(𝑥, 𝑦) | (𝑥 − 2) 2 + (𝑦 − 3) 2 = 4}. 𝑇 is an isometry
mapping point 𝐴(2,3) on 𝐴′(1,7).
Find the equation of the set 𝑇(𝐼).
Is the map of I also circle? Why?

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13. In the following figure, there are three points which are not collinear, it is 𝑃, 𝑄, 𝑅. 𝑇
and 𝑆 are isometry where 𝑃′ = 𝑇(𝑃), 𝑄′ = 𝑇(𝑄), 𝑅′ = 𝑇(𝑅), and 𝑃" = 𝑆(𝑃), 𝑄′ =
𝑆(𝑄), 𝑅′ = 𝑆(𝑅).
What kind of isometry 𝑇 and 𝑆 are those?
14. Isometry 𝑇 maps point 𝐴 to 𝑃, point 𝐵 to 𝑄 and 𝐶 to 𝑅. If 𝑇 is an opponent isometry,
find (sketching) the position of point 𝑅.
15. Find the coordinates of the point 𝑃 on the 𝑥 axis, measure of ∠𝐴𝑃𝑂 = ∠𝐵𝑃𝑋, if
𝐴 = (0,3) and 𝐵 = (6,5)
16. Suppose line g and points 𝐴, 𝐵 and 𝐴’ where 𝐴’ = 𝑀𝑔(𝐴), and 𝐴𝐵⃡⃗⃗⃗⃗ //𝑔. By using
only one ruler, determine the coordinate of point 𝐵′ = 𝑀𝑔(𝐵).

22
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:
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 𝐺(𝑍) = 𝑌, it is obtained that 𝐺[𝐹(𝑋)] = 𝑌 or (𝐺 ∙ 𝐹)(𝑋) = 𝑌, so 𝐻( 𝑋) = 𝑌 meaning
that 𝐻 is surjective.
• Is 𝐻 injective?
Take arbitrary 𝑃, 𝑄 elements of 𝑉 where 𝑃 ≠ 𝑄
Suppose ( 𝑃) = 𝐻( 𝑄) , then 𝐺[ 𝐹( 𝑃)] = 𝐺[ 𝐹( 𝑄)]
Since 𝐺 is injective then 𝐹(𝑃) = 𝐹(𝑄), and since 𝐹 injective then 𝑃 = 𝑄. It is contrary to which
it is known that 𝑃 ≠ 𝑄. It means that the assumption that 𝐻(𝑃) = 𝐻(𝑄) is not true. Therefore,
it should be 𝐻(𝑃) ≠ 𝐻(𝑄) implying 𝐻 is injective.
Since 𝐻 is surjective as well as injective, then 𝐻 is bijective. Thus, 𝐻 is a transformation.
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.

23
Theorem
The composition of two transformations is a transformation.
Example 1
Suppose that transformation 𝑇1[(𝑥, 𝑦)] = (𝑥 + 2, 𝑦) and 𝑇2[(𝑥, 2𝑦). If 𝑇 is the composition of
𝑇1 and 𝑇2, find 𝑇.
Solution:
𝑇 is the transformation 𝑇1 and 𝑇2 then :
𝑇[(𝑥, 𝑦)] = (𝑇1 ∙ 𝑇2 )(𝑥, 𝑦)
= 𝑇2[𝑇1 (𝑥, 𝑦)]
= 𝑇2 [(𝑥 + 2, −𝑦)]
= (𝑥 + 2, −2𝑦)
So the transformation 𝑇 is [(𝑥, 𝑦)] = (𝑥 + 2, −2𝑦) .
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.
B. Inverse Transformation
If Mg(P) = P′
, then Mg. Mg(P) = P or Mg
2
(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
Theorem
If T1, T2, and T3 are the transformation then T1[T2.T3]=[T1.T2].T3

24
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
Proof:
Suppose 𝑇 is a transformation. We define the equivalent 𝐹 as follows:
Suppose 𝑋 is an element of 𝑉, where 𝑉 is an euclidean field. Since 𝑇 is a transformation, then 𝑇
is bijective implying that A∈ 𝑉, such that 𝑇(𝐴) = 𝑋.
We define then 𝐹(𝑋) = 𝐴. meaning that 𝐹(𝑋) is the pre-image of 𝑋, such that from 𝑇(𝐴) = 𝑋,
𝑇[𝐹(𝑋)] = 𝑋 or (𝑇. 𝐹)(𝑋) = 𝐼(𝑋), for every 𝑋 element of 𝑉. So 𝐹𝑇 = 𝐼, so that 𝑇𝐹 = 𝐹𝑇 = 𝐼.
Now it shall be proved that L is a transformation. From this definition, it is clear that F is
surjective. Suppose 𝐹(𝑋1) = 𝐹(𝑋2) and suppose 𝑇(𝐴1) = 𝑋1, 𝑇(𝐴2) = 𝑋2 where 𝐹(𝑋1) = 𝐴1
and 𝐹(𝑋2) = 𝐴2
Since T is a transformation, then 𝐴1 = 𝐴2, and it is also obtained that 𝑋1 = 𝑋2 such that F is
injective. Thus, it proves that 𝐹 is injective.
So 𝐹 is a transformation.
The transformation 𝐹 is called the inverse transformation of 𝑇 and it is denoted by 𝐹 = 𝑇−1
.
Proof:
Suppose T is a transformation with two inverses, namely S1 and S2 then
( 𝑇 ∙ 𝑆1 )( 𝑃) = ( 𝑆1 ∙ 𝑇)( 𝑃) = 𝐼( 𝑃) for all 𝑃 and
Theorem
Each transformation 𝑇 has an inverse
Theorem
Transformation has exactly one inverse.

25
( 𝑇 ∙ 𝑆2 )( 𝑃) = ( 𝑆2 ∙ 𝑇)( 𝑃) = 𝐼( 𝑃) for all 𝑃
so, ( 𝑇 ∙ 𝑆1 )( 𝑃) = ( 𝑇 ∙ 𝑆2 )( 𝑃) ⇒ 𝑇 [ 𝑆1 ( 𝑃)] = 𝑇 [𝑆2 ( 𝑃)]
because 𝑇 is injective , then 𝑆1 ( 𝑃) = 𝑆2 ( 𝑃) for all 𝑃
So, 𝑆1 = 𝑆2 = 𝑆
Proof:
A reflection in the line 𝑔 is Mg
For X 𝜖 𝑔 then Mg (X) = X, So Mg . Mg (X) = Mg (X) = X = I(X)
So, Mg . Mg = I
Thus , the Mg−1
= 𝑀𝑔
For X ∉ 𝑔 , Mg (X) = X’, so the 𝑔 axis 𝑋𝑋̅̅̅̅ , then Mg . Mg (X) = Mg (X’) = X = I(X)
with 𝑔 axis 𝑋𝑋̅̅̅̅ . So, Mg . Mg = I atau Mg
-1
= Mg
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
Since every transformation has only one inverse, then (TOS) -1
= S-1
T-1
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
Theorem
Inverse of any reflection in a line is a reflection itself.
Definition
A transformation which has inverse which is the transformation itself is called an
involution.
Theorem
If 𝑇 and 𝑆 are transformations, then ( 𝑇 ∘ 𝑆) = 𝑆−1
∘ 𝑇−1

26
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
K

27
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 <RQR "when O is the origin
5. Let g is x-axis, and h = {(x, y | y = x}. S is a mapping defined as follows. If P∈g then S (P)
= P, And if P∈g then S (P) is the midpoint of the perpendicular line from P to G.
a. Prove that S is a transformation.
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).

28
c. S Mg.S. (A).
d. Mh.S .Mg (P).
e. S2
.Mh (P).
f. S Mg
2
(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.
a. Mg
3
(A)
b. Mh Mg Mh (A)
c. Mh Mg Mh Mh Mg (A)
d. Mg
2
Mh
3
(A)
8. Simplify.!
a. (WgVhMg)-1
b. (Mh Vh Wg Mg)-1
9. Suppose the transformations T1 [(x,y)] = (-x,y) and T2 [(x,y)]= (x,
1
2
y). Find the formula for
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!

29
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 𝐴 is a mapping 𝑆𝐴 that is defined for each point 𝑃 on a plane as follows :
i. If 𝑃 = 𝐴 , then 𝑆𝐴( 𝑃) = 𝑃.
ii. If 𝑃 ≠ 𝐴 , then𝑆𝐴( 𝑃) = 𝑃′, where 𝐴 as the center point of 𝑃𝑃′̅̅̅̅̅
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 𝐴(𝑎, 𝑏),𝑆𝐴 map point 𝑃(𝑥, 𝑦) to 𝑃′(𝑥′
, 𝑦′
), then𝑆𝐴( 𝑃) = 𝑃′ where 𝐴 is the center point
of 𝑃𝑃′̅̅̅̅̅ so
𝑥+𝑥′
2
= 𝑎 and
𝑦+𝑦′
2
= 𝑏.
It is obtained that 𝑥′
= −𝑥 + 2𝑎 and 𝑦′
= −𝑦 + 2𝑏.
Thus, if 𝐴 = (𝑎, 𝑏)and 𝑃 (𝑥, 𝑦), then 𝑆𝐴( 𝑃) = (2𝑎 − 𝑥, 2𝑏 − 𝑦).
Theorem
If 𝑔 and ℎ intersect and are perpendicular at point 𝐴, then 𝑆𝐴 = 𝑀𝑔 𝑀ℎ.

30
Proof :
Since 𝑔 ⊥ ℎ, we can make a system of orthogonal axis where 𝑔 as axis 𝑥 and ℎ as axis 𝑦, and 𝐴
is used as the point of origin. It must be proved that each point 𝑃 (𝑥, 𝑦), satisfies 𝑆𝐴(𝑃) =
𝑀𝑔 𝑀ℎ(𝑃).
Let 𝑃 (𝑥, 𝑦) ≠ 𝐴and 𝑆𝐴(𝑃) = 𝑃′′(𝑥1, 𝑦1).
Since 𝐴 is the center point 𝑃𝑃′′̅̅̅̅̅, so (0,0) = (
𝑥1+𝑥
2
,
𝑦1+𝑦
2
), so 𝑥1 = −𝑥 and 𝑦1 = −𝑦, therefore,
𝑆𝐴( 𝑃) = (−𝑥, −𝑦).
While (𝑀𝑔 𝑀ℎ)(𝑃) = 𝑀𝑔[ 𝑀ℎ(𝑃)] = 𝑀𝑔[(−𝑥, 𝑦)] = (−𝑥, −𝑦)
It turns out that 𝑆𝐴(𝑃) = (𝑀𝑔 𝑀ℎ)(𝑃) = (𝑥, 𝑦)
Thus, 𝑆𝐴 = 𝑀𝑔 𝑀ℎ
Theorem
If 𝑔 and ℎ are two lines perpendicular to each other, then 𝑀𝑔 𝑀ℎ = 𝑀ℎ 𝑀𝑔
Proof
If 𝑃 = 𝐴 (Look at the figure of the previous theorem), so 𝑀𝑔 𝑀ℎ( 𝑃) = 𝑀𝑔( 𝑃) = 𝑃. Whereas
𝑀ℎ 𝑀𝑔(𝑃) = 𝑀ℎ( 𝑃) = 𝑃, so 𝑀𝑔 𝑀ℎ(𝑃) = 𝑀ℎ 𝑀𝑔(𝑃). if 𝑃 ≠ 𝐴, then 𝑀𝑔 𝑀ℎ = 𝑆𝐴, while
𝑀ℎ 𝑀𝑔( 𝑃) = 𝑀ℎ[(𝑥, 𝑦)] = (−𝑥, −𝑦) = 𝑆𝐴(𝑃)
It turns out that 𝑀𝑔 𝑀ℎ = 𝑀ℎ 𝑀𝑔 = 𝑆𝐴
So,𝑀𝑔 𝑀ℎ = 𝑀ℎ 𝑀𝑔.

31
Theorem
If 𝑆𝐴 is halfturn, then 𝑆𝐴
−1
= 𝑆𝐴.
Proof:
Suppose that 𝑔 and ℎ are two lines perpendicular to each other and intersect at point 𝐴,
then 𝑀𝑔 𝑀ℎ = 𝑆𝐴, implying that 𝑆𝐴
−1
= (𝑀𝑔 𝑀ℎ)
−1
= 𝑀ℎ
−1
𝑀𝑔
−1
= 𝑀ℎ 𝑀𝑔 = 𝑀𝑔 𝑀ℎ = 𝑆𝐴. So
𝑆𝐴
−1
= 𝑆𝐴.
Reflection in line 𝑔 which is defined as 𝑀𝑔( 𝑃) = 𝑃 when 𝑃 ∈ 𝑔, and 𝑀𝑔( 𝑃) = 𝑃′
. with
𝑔 is the axis 𝑃𝑃̅̅̅̅ when the 𝑃 ∉ 𝑔. When we look in general, so for every 𝑃 ∈ 𝑔 the image of
point 𝑃 is the point itself. Such point is called invariant point of the reflection.
Definition:
Point 𝐴 is called the invariant transformation of T, if it is satisfied that 𝑇( 𝐴) = 𝐴
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 :
One example of collineation which is dilatation is halfturn. The example was verified
by the following theorem :
Proof :
Definition:
A collineation (∆) is called dilatation, if for every line 𝑔, it satisfies the property
∆𝑔//𝑔.
Theorem:
Suppose 𝑆𝐴 is a halfturn and 𝑔 is a line, if 𝐴 ∉ 𝑔, then 𝑆𝐴(𝑔)//𝑔

32
Suppose 𝑃 ∈ 𝑔 then 𝐴 is the midpoint of the segment 𝑃𝑃’̅̅̅̅̅ with 𝑃′
= 𝑆𝐴(𝑃).
Suppose 𝑄 ∈ 𝑔 then 𝐴is the midpoint of the segment 𝑄𝑄′̅̅̅̅̅ with 𝑄′
= 𝑆𝐴(𝑄).
Since ∆𝐴𝑃𝑄 ≅ ∆𝐴𝑃′𝑄′ then 𝑃𝑄𝑃′𝑄′ a parallelogram, implying that 𝑃𝑄 // 𝑃′𝑄". Thus, 𝑔 // 𝑆𝐴(𝑔).
Example 1 :
Suppose that two lines 𝑔 and ℎ are not parallel, 𝐴 is a point not located neither on 𝑔 nor ℎ.
Determine all points 𝑋 on 𝑔 and all points 𝑌 on ℎ such that 𝐴 is the midpoint of the segment 𝑋𝑌̅̅̅̅.
Solution :
Take a point 𝑃 ∈ 𝑔. Sketching 𝑃′
= 𝑆𝐴(𝑃). Then 𝑔′
= 𝑆𝐴(𝑔) passes through 𝑃′ where 𝑃𝐴 =
𝐴𝑃′, 𝑔′ // 𝑔. If 𝑔′ intersecting ℎ in 𝑌, then draw a line YA intersecting 𝑔 in 𝑋. Then 𝑋 and 𝑌 are
the pair of the points which seems to be exactly the only one pair
a. Prove that 𝑋 and 𝑌 is the only pair that satisfy the condition.
b. If we didn’t use 𝑔′
= 𝑆𝐴(𝑔) but we used ℎ′′
= 𝑆𝐴(ℎ), could we get another pair?
Proof :
Suppose 𝑃 is the center of a halfturn, and 𝑔 is a line. We must prove that:
a. 𝑆 𝑃(𝑔) // 𝑔
b. 𝑆 𝑃 𝑆 𝑃 = 𝐼, where 𝐼 is an identity transformation
Theorem :
A halfturn is a dilatation which has involutoric characteristic.

33
Prove:
a. It means that 𝑆 𝑃( 𝑔) = 𝑔′ is a line.
Suppose 𝐴 ∈ 𝑔, 𝐵 ∈ 𝑔, then 𝐴′
∈ 𝑔′
, and 𝑃𝐴 = 𝑃𝐴′
, 𝑃𝐵 = 𝑃𝐵′
, while 𝑚 (∠𝐴𝑃𝐵) =
𝑚(∠𝐴′
𝑃′
𝐵′
). Since ⊿ 𝑃𝐴𝐵 ≅ ⊿ 𝑃𝐴′𝐵′, 𝐴𝐵𝐴′𝐵′ is a parallelogram implying 𝑔′// 𝑔.
b. Since 𝑆 𝑃 𝑆 𝑃( 𝐴) = 𝑆 𝑃( 𝐴′) = 𝐴 for all points 𝐴 ∈ 𝑔, then 𝑆 𝑃 𝑆 𝑃( 𝑔) = 𝐼( 𝑔).
So 𝑆 𝑃 𝑆 𝑃 = 𝐼 meaning that 𝑆 𝑃 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.
Proof:
Suppose 𝐴 and 𝐵 are the centers of the halfturns. Suppose 𝑔 = 𝐴𝐵, both ℎ and 𝑘 are
perpendicular line to 𝐴𝐵 in 𝐴 and 𝐵 respectively, then:
𝑆𝐴 𝑆 𝐵 = (𝑀ℎ 𝑀𝑔)(𝑀𝑔 𝑀𝑘)
= [(𝑀ℎ 𝑀𝑔)𝑀𝑔]𝑀𝑘
= [𝑀ℎ (𝑀𝑔 𝑀𝑔)]𝑀𝑘
= 𝑀ℎ 𝐼 𝑀𝑘
= 𝑀ℎ 𝑀𝑘
Suppose that 𝑋 is the invariant point of SASB, then 𝑆𝐴 𝑆 𝐵(𝑋) = 𝑋 implying (𝑀ℎ 𝑀𝑘) (𝑋) = 𝑋 or
𝑀𝑘(𝑋) = 𝑀ℎ(𝑋).
Suppose 𝑀𝑘(𝑋) = 𝑋1
Theorem
The composition of two halfturns with different centers does not have invariant
point.

34
If 𝑋 ≠ 𝑋1, then each ℎ and 𝑘 is 𝑋𝑋1 axis and since a line segment has exactly one axis, it must
be ℎ = 𝑘. It is certainly not possible since 𝐴 ≠ 𝐵.
If 𝑋 = 𝑋1 then 𝑀𝑘(𝑋) = 𝑋1 and 𝑀ℎ(𝑋) = 𝑋1 , So 𝑋 ∈ ℎ and 𝑋 ∈ 𝑘 meaning that 𝑘 and ℎ
intersect at point 𝑋 which is not possible because ℎ // 𝑘.
Indeed, there can not be a point 𝑋 such that 𝑀 𝐾(𝑋) = 𝑀ℎ(𝑋) or 𝑆𝐴 𝑆 𝐵 = 𝑋.
So 𝑆𝐴 𝑆 𝐵 doesn’t have a fixed point.
Proof:
Suppose there are two halfturns 𝑆 𝐷 and 𝑆 𝐸, so 𝑆 𝐷(𝐴) = 𝐵 and 𝑆 𝐸 (𝐵) = 𝐴.
So, 𝑆 𝐷( 𝐴) = 𝑆 𝐸 (𝐵), then 𝑆 𝐷[𝑆 𝐷(𝐴)] = 𝑆 𝐷[𝑆 𝐸(𝐴)], then 𝐴 = 𝑆 𝐷 𝑆 𝐸(𝐴). If 𝐷 and 𝐸 are
two different points, it means that 𝐴 is a fixed point 𝑆 𝐷 𝑆 𝐸. 𝐼 which is not possible. Therefore,
there is no more than one halfturns that maps 𝐴 to 𝐵. The only halfturn is 𝑆 𝑇(𝐴) = 𝐵 where 𝑇 is
the midpoint of 𝐴𝐵̅̅̅̅.
Example:
Given 𝐸 = {( 𝑥, 𝑦)|𝑥2
+ 4𝑦2
= 16}, 𝐴(4, −3), and 𝐵(3,1).If 𝑔 is the 𝑋 axis, show that 𝐴 ∈
𝑀𝑔 𝑆 𝐵(𝐸)?
Solution:
It is known that (𝑀 𝑔S 𝐵)−1
= S 𝐵
−1
. M 𝑔
−1
= S 𝐵 𝑀𝑔
If 𝑃(𝑥, 𝑦), then 𝑀𝑔( 𝑃) = (𝑥, −𝑦) then S 𝐵( 𝑃) = (2.3 − 𝑥, 2.1 − 𝑦) = (6 − 𝑥, 2 − 𝑦)
𝐴 ∈ 𝑀𝑔 𝑆 𝐵(𝐸) ⇔ 𝑆 𝐵 𝑀𝑔(𝐴) ∈ 𝐸
𝑆 𝐵 𝑀𝑔( 𝐴) = 𝑆 𝐵[𝑀𝑔(4, −3)] = 𝑆 𝐵(4,3) = (2, −1)
Since (2, −1) ∉ 𝐸, then (𝑀𝑔 𝑆 𝐵)
−1
( 𝐴) ∉ ( 𝐸)
In a similar way, we can define a set of maps if the equations are known.
In the latest example we know that 𝑃 ∈ 𝑀𝑔 𝑆 𝐵( 𝐸) if and only if (𝑀𝑔 𝑆 𝐵)
−1
( 𝑃) ∈ ( 𝐸).
If 𝑃(𝑥, 𝑦) so (𝑀𝑔 𝑆 𝐵)
−1
( 𝑃) = (6 − 𝑥, 2 + 𝑦), then (𝑀𝑔 𝑆 𝐵)
−1
( 𝑃) ∈ ( 𝐸) if and only if
(6 − 𝑥, 2 + 𝑦) ∈ {( 𝑥, 𝑦)|𝑥2
+ 4𝑦2
= 16}.
Theorem
If 𝐴 and 𝐵 are two different points, then there is only one and a halfturn that
maps 𝐴 to 𝐵.

35
So it must be (6 − 𝑥)2
+ 4(2 + 𝑦)2
= 16. 𝑃( 𝑥, 𝑦) ∈ 𝑀𝑔 𝑆 𝐵( 𝐸) if and only if 𝑃( 𝑥, 𝑦) ∈
{( 𝑥, 𝑦)|𝑥2
+ 4𝑦2
− 12𝑥 + 16𝑦 + 36 = 0}
Therefore 𝑥2
+ 4𝑦2
− 12𝑥 + 16𝑦 + 36 = 0 is the equation of the image of 𝐸 by
transformation of 𝑀𝑔 𝑆 𝐵.
Exercises
1. Suppose three distinct points A, B, P are not collinear, sketch
a. 𝑆𝐴(𝑃)
b. 𝑆𝐴 𝑆 𝐵( 𝑃)
c. R so 𝑆 𝐵( 𝑅) = 𝑃
d. 𝑆𝐴
2
( 𝑃)
2. Given the line g and point A, 𝐴 ∉ 𝑔
a. Draw lines 𝑔′
= 𝑆𝐴( 𝑔), why 𝑆𝐴( 𝑔) is a line?
b. Prove that 𝑔′
// 𝑔
3. Given ∆ABC and a parallelogram WXYZ. There is a point K which lies outside the triangle
∆ABC and the parallelogram WXYZ.
a. Sketh 𝑆K(∆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)
c. S A
-1
(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)
c. (M g S c)
-1
(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)
c. S A (E) = E
b. S A (D) = S B (D)
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)

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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 𝐴𝑇̅̅̅̅, so that the X ∈ C, Y ∈ g such
that the P midpoint of 𝑋𝑌̅̅̅̅.
b. Prove that the construction is correct.

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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
The notation of vector
- The notation of vector 𝐴𝐵 is 𝐴𝐵⃗⃗⃗⃗⃗
- The notation of vector 𝐶𝐷 is 𝐶𝐷⃗⃗⃗⃗⃗
Vector𝐴𝐵 and 𝐶𝐷 are vector of which the points 𝐴 and 𝐶 are the the initial
points, and the points 𝐵 and 𝐷 are the terminal points.
As the illustration, look at the following figure:
The definition above can be interpreted also that 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ , if 𝑆 𝑝(𝐶) = 𝐵 and
𝑃 is the midpoint of 𝐴𝐷⃗⃗⃗⃗⃗ , as an illustration, look at the following figure:
Definition
Vector is a line segment of which one of the edge is the initial point and
the other edge is the terminal point.
Definition
𝐴𝐵⃗⃗⃗⃗⃗ is equivalent to 𝐶𝐷⃗⃗⃗⃗⃗ that is notated as 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ , if there is a halfturn
𝑆 𝑝( 𝐴) = 𝐷 where 𝑃 is the midpoint 𝐵𝐶⃗⃗⃗⃗⃗

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The equivalence of two vectors satisfies a characteristic asserted by the
following theorem
Proof:
Look at the latest figure:
1. Suppose that 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ , if 𝑃 is the midpoint of 𝐵𝐶⃗⃗⃗⃗⃗ , then 𝑆 𝑝(𝐴) = 𝐷, by the
definition of equivalence. The diagonals of quadrilateral 𝐴𝐵𝐶𝐷 are bisected
in 𝑃. Then 𝐴𝐵𝐶𝐷 is a parallelogram.
2. Suppose that 𝐴𝐵𝐶𝐷 is a parallelogram, then diagonals 𝐴𝐷 and 𝐵𝐶 are
intersected in the midpoint 𝑃 implying that 𝑆 𝑝(𝐴) = 𝐷. In other words, 𝑃 is
the midpoint of both 𝐴𝐷 and 𝐵𝐶.
So 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗
Consequence:
If 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ , then 𝐴𝐵𝐶𝐷, 𝐴𝐵⃗⃗⃗⃗⃗ and 𝐶𝐷⃗⃗⃗⃗⃗ are either parallel or colinear
Theorem
If there are two vectors 𝐴𝐵⃗⃗⃗⃗⃗ and 𝐶𝐷⃗⃗⃗⃗⃗ which are not collinear then 𝐴𝐵𝐶𝐷 is
a parallelogram if and only if 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗
Theorem
Suppose 𝐴𝐵⃗⃗⃗⃗⃗ , 𝐶𝐷⃗⃗⃗⃗⃗ dan 𝐸𝐹⃗⃗⃗⃗⃗ are vectors, then the following properties
apply:s
• 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ ( 𝑟𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒)
• If 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ then 𝐶𝐷⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ (𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐)
• If 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ then 𝐶𝐷⃗⃗⃗⃗⃗ = 𝐸𝐹⃗⃗⃗⃗⃗ then 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐸𝐹⃗⃗⃗⃗⃗ (𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒) are colinear.
Theorem
If 𝐴𝐵⃗⃗⃗⃗⃗ is a vector and there is a point 𝑃, then there is a unique point Q
such that 𝑃𝑄⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ .

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Proof :
Suppose R is the midpoint of 𝐵𝑃⃗⃗⃗⃗⃗ , 𝑄 = 𝑆 𝑅( 𝐴) then 𝐴𝐵⃗⃗⃗⃗⃗ = 𝑃𝑄⃗⃗⃗⃗⃗ 𝑜𝑟 𝑃𝑄⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ .
To prove the uniqueness of point, let 𝐴𝐵⃗⃗⃗⃗⃗ = 𝑃𝑇⃗⃗⃗⃗⃗ .
Then 𝑆 𝑅( 𝐴) = 𝑇, because R is thee midpoint of 𝐵𝑃⃗⃗⃗⃗⃗ and the image of 𝐴 by 𝑆 𝑅,
then 𝑇 = 𝑄 meaning that 𝑃𝑄⃗⃗⃗⃗⃗ is the only vector where 𝑃 is the initial point and
𝑄 is the terminal point which is equivalent to 𝐴𝐵⃗⃗⃗⃗⃗
Corollary 1:
If 𝑃1( 𝑥1, 𝑦1), 𝑃2( 𝑥2, 𝑦2), and 𝑃3( 𝑥3, 𝑦3) are points with the specified
coordinates, then 𝑃(𝑥3 + 𝑥2 − 𝑥1 , 𝑦3 + 𝑦2 − 𝑦1) is the only point that satisfies
𝑃3 𝑃⃗⃗⃗⃗⃗⃗⃗ = 𝑃1 𝑃2
⃗⃗⃗⃗⃗⃗⃗⃗
Corollary 2:
If 𝑃𝑛 = ( 𝑥 𝑛, 𝑦 𝑛), 𝑛 = 1,2,3,4 then 𝑃1 𝑃2
⃗⃗⃗⃗⃗⃗⃗⃗ = 𝑃3 𝑃4
⃗⃗⃗⃗⃗⃗⃗⃗ . If and only if 𝑥2 − 𝑥1 =
𝑥4 − 𝑥3, and 𝑦2 − 𝑦1 = 𝑦4 − 𝑦3.
The last concept that is related to the vector is a scalar multiplication by
vector. Suppose that 𝐴𝐵⃗⃗⃗⃗⃗ is vector and 𝑘 is real number, so:
a. If 𝑘 > 0, then 𝑘𝐴𝐵⃗⃗⃗⃗⃗ is a vector 𝐴𝑃⃗⃗⃗⃗⃗ which is defined as 𝐴𝑃⃗⃗⃗⃗⃗ = 𝑘(𝐴𝐵⃗⃗⃗⃗⃗ ), 𝑃 ∈
𝐴𝐵⃗⃗⃗⃗⃗ (ray 𝐴𝐵).
b. If 𝑘 < 0, then 𝑘𝐴𝐵⃗⃗⃗⃗⃗ where 𝑃 is the opposite ray of 𝐴𝐵⃗⃗⃗⃗⃗ and 𝐴𝑃⃗⃗⃗⃗⃗ = | 𝑘|(𝐴𝐵⃗⃗⃗⃗⃗⃗⃗ )

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B. Translation
In this section, the concept of translation is introduced using the definition of
vector.
Proof.
Choose a coordinate system with 𝑡 as the 𝑦-axis and a line which is
perpendicular to 𝑡 as the 𝑥-axis, and s which its parallel to t.
Suppose 𝐴 = ( 𝑎1, 𝑎2) and 𝐵 = ( 𝑏1, 𝑏2).
If 𝑁 𝑖𝑠 the midpoint of 𝐴′′𝐵⃗⃗⃗⃗⃗⃗⃗⃗ , then we have to prove 𝑆 𝑁( 𝐴) = 𝐵′′
.
If the equation of 𝑠 is 𝑥 = 𝑘( 𝑘 ≠ 0), then
𝐴′′
= 𝑀𝑠 𝑀𝑡( 𝐴) = 𝑀𝑠(−𝑎1, 𝑎2) = (2𝑘 + 𝑎1, 𝑎2)
𝐵′′
= 𝑀𝑠 𝑀𝑡( 𝐵) = 𝑀𝑠(−𝑏1, 𝑏2) = (2𝑘 + 𝑏1, 𝑏2)
Because we know that 𝑁 is the midpoint of 𝐴′′𝐵⃗⃗⃗⃗⃗⃗⃗⃗ , then
𝑁 = [
(2𝑘 + 𝑎1) + 𝑏1
2
,
𝑎2 + 𝑏2
2
] , whereas
𝑆 𝑁 = [2 [
2𝑘 + 𝑎1 + 𝑏1
2
] − 𝑎1, 2 [
𝑎2 + 𝑏2
2
] − 𝑎2] = (2𝑘 + 𝑏1, 𝑏2)
Evidently 𝑆 𝑁( 𝐴) = 𝐵′′
. Therefore 𝐴𝐴′′⃗⃗⃗⃗⃗⃗⃗⃗ = 𝐵𝐵′′⃗⃗⃗⃗⃗⃗⃗⃗
Theorem
If 𝑠 and 𝑡 are two parallel lines, then 𝐴 and 𝐵 are the two points, so 𝐴𝐴′′⃗⃗⃗⃗⃗⃗⃗⃗ =
𝐵𝐵′′⃗⃗⃗⃗⃗⃗⃗⃗ with 𝐴′′
= 𝑀𝑡 𝑀 𝑠(𝐴) and 𝐵′′
= 𝑀𝑡 𝑀𝑠(𝐵).

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In the figure above, it is clear that every vector determines a translation.
If𝐴𝐵⃗⃗⃗⃗⃗ is a vector then 𝐺 𝐴𝐵 is a symbol to address a translation in the length of 𝐴𝐵.
Proof:
If 𝑃 is an arbitrary point, it must be proved that 𝐺 𝐴𝐵( 𝑃) = 𝐺 𝐶𝐷( 𝑃). Suppose
𝐺 𝐴𝐵( 𝑃) = 𝑃1 and 𝐺 𝐶𝐷( 𝑃) = 𝑃2, so 𝑃𝑃1 = 𝐴𝐵⃗⃗⃗⃗⃗ and 𝑃𝑃2 = 𝐶𝐷⃗⃗⃗⃗⃗ . Since 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ , 𝑃𝑃1
⃗⃗⃗⃗⃗⃗⃗ =
𝑃𝑃2
⃗⃗⃗⃗⃗⃗⃗ meaning that 𝑃1 = 𝑃2 and 𝐺 𝐴𝐵 = 𝐺 𝐶𝐷
Proof:
Suppose 𝑃 is an arbitrary point, if 𝑃’ = 𝐺 𝐴𝐵 (𝑃) and 𝑃’’ = 𝑀𝑠 𝑀𝑡 (𝑃), so it is
necessary to prove that 𝑃’ = 𝑃’’
Definition
A mapping 𝐺 is a translation, if there is a vector 𝐴𝐵⃗⃗⃗⃗⃗ such that for every point
𝑃 in a plane 𝑉 has image 𝑃′
𝑤here 𝐺( 𝑃) = 𝑃′
and 𝑃𝑃′⃗⃗⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ .
Theorem
If 𝐴𝐵 = 𝐶𝐷 then 𝐺 𝐴𝐵 = 𝐺 𝐶𝐷
Theorem
Suppose 𝑡 and 𝑠 as the two lines is parallel and 𝐶𝐷⃗⃗⃗⃗⃗ is the vector that
perpendicular to 𝑠 and 𝑡, with 𝐶 ∈ 𝑠 and 𝐷 ∈ 𝑠. If 𝐴𝐵⃗⃗⃗⃗⃗ = 2𝐶𝐷⃗⃗⃗⃗⃗ so 𝐺 𝐴𝐵 = 𝑀𝑠 𝑀𝑡

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As the property of translation suggests, if 𝐺 𝐴𝐵(𝑃) = 𝑃’ then 𝑃𝑃’⃗⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ . Since,
𝐴𝐵⃗⃗⃗⃗⃗ = 2𝐶𝐷⃗⃗⃗⃗⃗ , 𝑃𝑃’⃗⃗⃗⃗⃗⃗ = 2𝐶𝐷⃗⃗⃗⃗⃗ .
Related to 𝐶’’ = 𝑀𝑠 𝑀𝑡( 𝐶), 𝐶 ∈ 𝑡 so 𝐶’’ = 𝑀𝑡(𝐶). It means that 𝐷 is the
midpoint of 𝐶𝐶’’⃗⃗⃗⃗⃗⃗⃗ implying that 𝐶𝐶’’⃗⃗⃗⃗⃗⃗⃗ = 2 𝐶𝐷⃗⃗⃗⃗⃗ . Since 𝐶𝐶’’⃗⃗⃗⃗⃗⃗⃗ = 𝑃𝑃’’⃗⃗⃗⃗⃗⃗⃗ , 𝑃𝑃’’⃗⃗⃗⃗⃗⃗⃗ = 2 𝐶𝐷⃗⃗⃗⃗⃗ = 𝑃𝑃’⃗⃗⃗⃗⃗⃗ ,
and it means that 𝑃’ = 𝑃’’, so 𝐺 𝐴𝐵 = 𝑀𝑠 𝑀𝑡.
Notes
1) Each translation 𝐺 𝐴𝐵 can be written as the composition between two reflections
in two lines which are perpendicular to 𝐴𝐵̅̅̅̅ and it is ½ 𝐴𝐵 in length.
2) If 𝐴𝐵 is a line and 𝐶 is the midpoint of 𝐴𝐵̅̅̅̅ whereas 𝑡, 𝑠 and 𝑛 are the perpendicular
lines to 𝐴𝐵 in 𝐴, 𝐶 and 𝐵 respectively, then 𝐺 𝑎𝑏 = 𝑀𝑠 𝑀𝑡 = 𝑀 𝑛 𝑀𝑠
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
Proof:
According to latest figure, 𝐺 𝐴𝐵 = 𝑀𝑠 𝑀𝑡 = 𝑀 𝑛 𝑀𝑠
While 𝐺 𝐵𝐴 = 𝑀𝑡 𝑀𝑠 = 𝑀𝑠 𝑀 𝑛
( 𝐺 𝐴𝐵)−1
= ( 𝑀𝑠 𝑀𝑡)−1
= 𝑀𝑡
−1
. 𝑀𝑠−1
= 𝑀𝑡 𝑀𝑠 = 𝐺 𝐵𝐴
Theorem
If 𝐺 𝐴𝐵 is a translation, then ( 𝐺 𝐴𝐵)−1
= 𝐺 𝐵𝐴

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So, ( 𝐺 𝐴𝐵)−1
= 𝐺 𝐵𝐴.
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:
Proof :
Suppose that 𝑔 = 𝐶𝐷⃗⃗⃗⃗⃗ , 𝑘 ⊥ 𝑔 𝑖𝑛 𝐶, 𝑚 ⊥ 𝑔 𝑖𝑛 𝐷
𝐴𝐵 is a vector from 𝑘 to 𝑚, therefore 𝐴𝐵⃗⃗⃗⃗⃗ = 2𝐶𝐷⃗⃗⃗⃗⃗ , so 𝐺 𝐴𝐵 = 𝑆 𝐷 𝑆 𝐶. While 𝑆 𝐷 =
𝑀 𝑚 𝑀𝑔 and 𝑆𝑐 = 𝑀𝑔 𝑀𝑘.
Then 𝑆 𝐷 𝑆 𝐶 = (𝑀 𝑚 𝑀𝑔)(𝑀𝑔 𝑀𝑘) = 𝑀 𝑚(𝑀𝑔 𝑀𝑔)𝑀𝑘 = 𝑀 𝑚 𝑀𝑘. 𝑆𝑜 𝐺 𝐴𝐵 = 𝑆 𝐷 𝑆 𝐶.
Example 1
Given 𝐴 = (3, −1), 𝐵 = (1,7) and 𝐶 = (4,2). Find the coordinate of a point
𝐷 such that 𝐺 𝐴𝐵 = 𝑆 𝐷 𝑆 𝐶.
Solution:
Suppose that 𝐸 is a point such that 𝐶𝐸 = 𝐴𝐵, then 𝐸[4 + (1 − 3),2 + (7 − (−1))]
or
𝐸 = (2,10). If 𝐷 is the midpoint 𝐶𝐸̅̅̅̅, then 𝐷 = (3,6), implying that 𝐶𝐸⃗⃗⃗⃗⃗ = 2𝐶𝐷⃗⃗⃗⃗⃗ .
Theorem
If 𝐺 𝐴𝐵 is a translation, C and D are points such that 𝐴𝐵⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ , then 𝐺 𝐴𝐵 = 𝑆 𝑝 𝑆 𝐶.

44
Thus, 𝐴𝐵⃗⃗⃗⃗⃗ = 2𝐶𝐷⃗⃗⃗⃗⃗ , which obtains 𝐺 𝐴𝐵 = 𝑆 𝐷 𝑆 𝐶 where 𝐷 (3,6).
Proof:
Suppose that 𝐺 𝐴𝐵 is a translation and 𝐶 is any point and suppose that 𝐸 is also a point
so
𝐶𝐸⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ . If 𝐷 mid point 𝐶𝐸̅̅̅̅̅then 𝐶𝐸⃗⃗⃗⃗⃗ = 2𝐶𝐷⃗⃗⃗⃗⃗ .
According to previous theorem 𝐺 𝐴𝐵 = 𝑆 𝐷 𝑆 𝐶, then 𝐺 𝐴𝐵 𝑆 𝐶 = 𝑆 𝐷 𝑆 𝐶 𝑆 𝐶 = 𝑆 𝐷 𝐼 =
𝑆 𝐷. So 𝐺 𝐴𝐵 𝑆 𝐶 = 𝑆 𝐷.
As a result of the theorem above is:
If 𝑆𝐴, 𝑆 𝐵 and 𝑆 𝐶 is half round, then 𝑆𝐴 𝑆 𝐵 𝑆 𝐶 = 𝑆 𝐷, with 𝐷 is points that satisfy 𝐴𝐷⃗⃗⃗⃗⃗ =
𝐵𝐶⃗⃗⃗⃗⃗ .
To declare the composition of two translations is a translation in Cartesian
coordinates, consider the following theorem.
Proof:
For 𝑃(𝑥, 𝑦), 𝑇(𝑃) = (𝑥 + 𝑎, 𝑦 + 𝑏). Suppose that 𝑃’ = 𝐺 𝑂𝐴(𝑃) then 𝑃𝑃’ = 𝑂𝐴
implying that 𝑃’ = (𝑥 + 𝑎 − 0, 𝑦 + 𝑏 − 0) = (𝑥 + 𝑎, 𝑦 + 𝑏). Thus 𝑇(𝑃) =
𝐺 𝑂𝐴(𝑃) for each 𝑃 ∈ 𝑉. In other words 𝐺 𝑂𝐴 = 𝑇.
Example 2
Suppose that 𝐺 𝐴𝐵 is a translation which takes point 𝐴(2,3) to point 𝐵(4,1) and 𝐺 𝐶𝐷
is a translation which takes point 𝐶(−3,4) to point 𝐵(0,3). If 𝑃(𝑥, 𝑦). Determine the
coodinate of 𝐺 𝐶𝐷 𝐺 𝐴𝐵(𝑃).
Theorem
The composition of a translation and a halfturn is a halfturn.
Theorem
If 𝐺 𝑂𝐴 is a translation where the coordinates of point 0 and point 𝐴 are respectively
(0,0) and (𝑎, 𝑏). 𝑇 is a transformation that maps each point 𝑃 (𝑥, 𝑦) to 𝑇 (𝑃) =
(𝑥 + 𝑎, 𝑦 + 𝑏) then 𝐺 𝑂𝐴 = 𝑇.

45
Solution:
Suppose that 0’ = 𝐺 𝐴𝐵(0) and 0’’ = 𝐺 𝐶𝐷(𝑂) then 𝑂𝑂′⃗⃗⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ and 𝑂𝑂′′⃗⃗⃗⃗⃗⃗⃗⃗ = 𝐶𝐷⃗⃗⃗⃗⃗ .
Then 0’ = (0 + 4 − 2,0 + 1 − 3) = (2, −2) and
0’’ = (0 + 0 + 3,0 + 3 − 4) = (3, −1).
So 𝐺 𝐴𝐵(𝑃) = (𝑥 + 2, 𝑦 − 2) and 𝐺 𝐶𝐷(𝑃) = (𝑥 + 3, 𝑦 − 1).
Thus 𝐺 𝐶𝐷 𝐺 𝐴𝐵( 𝑃) = 𝐺 𝐶𝐷[( 𝑥 ± 2, 𝑦 − 2)]
= ( 𝑥 + 2 + 3, 𝑦 − 2 − 1)
= (𝑥 + 5, 𝑦 − 3)
Exercises
1. Suppose there are 4 points that are wirtten as 𝐴, 𝐵, 𝐶 and 𝐷 of which each pair of
three points is not collinear. Skecth the following:
a. Point E such that 𝐶𝐸⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗
b. Point F such that 𝐷𝐹⃗⃗⃗⃗⃗ = 𝐵𝐴⃗⃗⃗⃗⃗
c. 𝑆𝐴(𝐴𝐵)
2. Given points 𝐴(0,0), 𝐵(5,3), and 𝐶(−2,4). Find the coordinate :
a. R so that such that 𝐴𝑅⃗⃗⃗⃗⃗ = 𝐵𝐶⃗⃗⃗⃗⃗
b. S so that such that 𝐶𝑆⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗
c. T so that such that 𝑇𝐵⃗⃗⃗⃗⃗ = 𝐴𝐶⃗⃗⃗⃗⃗
3. If 𝐴(1,3), 𝐵(2,7), dan 𝐶(−1,4) are the vertices of parallelogram 𝐴𝐵𝐶𝐷. Determine
the coordinate of the point 𝐷.
4. Suppose there are two lines 𝑔 and ℎ whcih are parallel, point 𝑃 ∈ 𝑔, and point 𝑄
neither on 𝑔 nor on ℎ.
a. Sketch 𝑃’ = 𝑀ℎ 𝑀𝑔(𝑃) and 𝑄’ = 𝑀𝑔 𝑀ℎ(𝑄)
b. Prove that 𝑃𝑃′⃗⃗⃗⃗⃗⃗ = 𝑄𝑄′⃗⃗⃗⃗⃗⃗⃗
5. Given a line 𝑔 and circles 𝐿1 and 𝐿2. Line 𝑔 are not cuting the circles 𝐿1 and 𝐿2. Use
a transformation to draw a square having vertices located on 𝑔, a vertex that is located
on 𝐿1and the other vertex points located on 𝐿2

46
6. If 𝑃0 = (0,0), 𝑃1 = (𝑥1, 𝑦1), 𝑃2 = (𝑥2, 𝑦2), and 𝑃3 = (𝑥3, 𝑦3) while 𝑘 > 0.
a. Find the coordinate of 𝑃 such that 𝑃0 𝑃1
⃗⃗⃗⃗⃗⃗⃗⃗ = 𝑘𝑃0 𝑃1
⃗⃗⃗⃗⃗⃗⃗⃗
b. Find the coordinate of 𝑃 such that 𝑃1 𝑃⃗⃗⃗⃗⃗⃗⃗ = 𝑘𝑃1 𝑃2
⃗⃗⃗⃗⃗⃗⃗⃗
c. Is the statement “If 𝑃3 𝑃⃗⃗⃗⃗⃗⃗⃗ = 𝑘𝑃1 𝑃2
⃗⃗⃗⃗⃗⃗⃗⃗ , then 𝑃 = {𝑥3 + 𝑘(𝑥2 − 𝑥1), 𝑦3 + 𝑘(𝑦2 −
𝑦1)}.” valid for 𝑘 < 0?
7. Given 𝐴, 𝐵, and 𝐶 which are not collinear. Sketch:
a. 𝐺 𝐴𝐵(𝐴) and 𝐺 𝐴𝐵(𝐵).
b. 𝐺 𝐴𝐵(𝐶).
c. The lines 𝑔 and ℎ where 𝐴 ∈ 𝑔 and 𝐺 𝐴𝐵 = 𝑀ℎ 𝑀𝑔.
8. Suppose there are two points that notated as 𝐴 and 𝐵 and line 𝑔 such that 𝑔 ⊥ 𝐴𝐵⃗⃗⃗⃗⃗ .
Sketch:
a. Line ℎ such that 𝑀ℎ 𝑀𝑔 = 𝐺 𝐴𝐵.
b. Line 𝑘 such that 𝑀𝑔 𝑀𝑘 = 𝐺 𝐴𝐵.
c. Line 𝑚 such that 𝑚’ = 𝐺 𝐴𝐵(𝑚).
d. Point 𝐶 such that the 𝐺 𝐵𝐴 (𝐶) = 𝐵
9. Suppose lines 𝑔 and ℎ are parallel and there is point 𝐴 which is not on the line.
a. Draw the point 𝐵 such that 𝑀ℎ 𝑀𝑔 = 𝐺 𝐴𝐵
b. Draw the point 𝐶 such that 𝑀𝑔 𝑀ℎ = 𝐺2𝐴𝐶
10. Given 𝐴 (2,3) and 𝐵 (−4,7), determine the equation of a line 𝑔 and ℎ such that
𝑀ℎ 𝑀𝑔 = 𝐺 𝐴𝐵
11. Given three points 𝐴 (−1,3), 𝐵 (5, −1) and 𝐶 (2,4)
a. Find the coordinate 𝐶 ′ = 𝐺 𝐴𝐵 (𝐶)
b. Find the equation of lines 𝑔 and ℎ so 𝐶 ∈ 𝑔 and so 𝑀ℎ 𝑀𝑔 = 𝐺 𝐴𝐵

47
12. The edges of a river are depicted with two parallel lines that is written as 𝑡 and 𝑠
(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 𝐷 to
the town 𝐸 will be as short as possible.

48
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
To symbolize an angle, for example ∠ ABC is a directed angle where 𝐵𝐴⃗⃗⃗⃗⃗ rays is the
initial ray and 𝐵𝐶⃗⃗⃗⃗⃗ is the terminal ray notated as ∠ ABC
For another purpose, <ABC cannot be written as <BCA. For directed angle <CBA, the
initial ray is 𝐵𝐶⃗⃗⃗⃗⃗ and the end ray is 𝐵𝐴⃗⃗⃗⃗⃗ .
In Euclidean geometry, it has been studied about the magnitude of an angle, i.e. every
angle ABC has magnitude in the interval from 0°
to 180°
, which is notated by 0°
≤ 𝑚 <
𝐴𝐵𝐶 ≤ 180°
.
It is the same as a directed angle < 𝐴𝐵𝐶 which always has a magnitude, however, it has
following properties:
a) If the triple orientation (𝐵𝐶𝐴) is positive, then 𝑚 < 𝐴𝐵𝐶 = 𝑚 < 𝐴𝐵𝐶
b) If the triple orientation (𝐵𝐶𝐴) is negative, then 𝑚 < 𝐴𝐵𝐶 = −(𝑚 < 𝐴𝐵𝐶)
Definition
Directed angle is an angle of which one of the ray is the initial ray and the other
is the terminal ray

49
When < 𝐴𝐵𝐶 is an angle then < 𝐴𝐵𝐶 = < 𝐶𝐵𝐴, so 𝑚 < 𝐴𝐵𝐶 = −𝑚 < 𝐶𝐵𝐴.
But for a directed angle < 𝐴𝐵𝐶 it applies for the 𝑚 < 𝐴𝐵𝐶 =
−𝑚 𝐶𝐵𝐴 since the orientation of the 𝐵𝐴𝐶 is always contrary to the orientation of the 𝐵𝐶𝐴.
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.

50
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
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)

51
RA, α(P)
PA
RA, -α(P)
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.
Based on the definition, the rotation 𝑅 𝐴,𝛼, just has an invarioant point, i.e. A (the
center of the rotation). The image of point the P by rotation 𝑅 𝐴,𝛼, 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.
Defenition
If A is a point and 𝛼 is angle where -180°<𝛼<180°, a rotation with center A and
angle 𝛼 denoted by 𝑅 𝐴,𝛼 is a function from V to V defined by
1. If P = A, so 𝑅 𝐴,𝛼(P) = P.
2. If P ≠ A, so 𝑅 𝐴,𝛼(P) = P’, then m (< PAP”) = 𝛼and AP’=AP.
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/2 , then RA,α =MtMs

52
A P
u
s
t
v
½ α
½ α
K
t
s
½ α
Proof:
Suppose that a point K ≠ A located on s. If K’ = MtMs(K) so m(<KAK’) = 2. ½ α = α.
Since <KAK’=α, so RA,α =MtMs.
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
1. Suppose that s = 𝐴𝑃̅̅̅̅, t is the bisecting line of <PAP’, and suppose the magnitude of
angle from s to t is ½ α, so RA,α = MtMs.
K
A
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

53
2. Suppose u = 𝐴𝑃′̅̅̅̅̅, 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, 𝛼1 andRB, 𝛼2 is two rotation, then RB, 𝛼2RA, 𝛼1 = Rc, 𝛼 with the conditions of 𝛼 as
follows:
1. If 0° < | 𝛼1 + 𝛼2| < 180° 𝑡ℎ𝑒𝑛 𝛼 = 𝛼1 + 𝛼2.
2. If 𝛼1 + 𝛼2 > 180° 𝑡ℎ𝑒𝑛 𝛼 = ( 𝛼1 + 𝛼2) − 360°.
3. If 𝛼1 + 𝛼2 < −180° 𝑡ℎ𝑒𝑛 𝛼 = ( 𝛼1 + 𝛼2) + 360°.
Example
RA,120 . RA,30= RA,150
RB,160 . RB,40= RB,-160 .
RC,-150 .RC,-50 .= RC,160 .
While if𝛼1 + 𝛼2 = 0°, 𝑡ℎ𝑒𝑛 RB, 𝛼2RA, 𝛼1is a rotation .
Theorem
The composition of two rotations is a rotation or translation.

54
Proof :
Assume there are rotations 𝑅 𝐴,𝛼and 𝑅 𝐵,𝛽. Draw a line s = 𝐴𝐵⃡⃗⃗⃗⃗ , if m(∠XAY) = m(∠XAZ)
=
1
2
𝛼, then 𝑅 𝐴,𝛼 = 𝑀𝑠 𝑀𝑡 and 𝑅 𝐵,𝛽 = 𝑀 𝑢 𝑀𝑠.
So 𝑅 𝐵,𝛽 𝑅 𝐴,𝛼 = 𝑀 𝑢 𝑀𝑠 𝑀𝑠 𝑀𝑡= 𝑀 𝑢 𝑀𝑡.
If u is parallel to t then 𝑅 𝐵,𝛽 𝑅 𝐴,𝛼 is a translation, and if u and t intersect at C, then
𝑀 𝑢 𝑀𝑡 is a rotation centered at C. Assume 𝑅 𝐶,0 = 𝑅 𝐵,𝛽 𝑅 𝐴,𝛼then in 𝛼, 𝛽, dan 𝜃, there is
following relationship:
m(∠ABC) =
1
2
𝛼, m(∠BAC) =
1
2
𝛽.
Therefore m(∠PCB) =
1
2
𝛼 +
1
2
𝛽. It means that the angle from t to u is
1
2
𝛼 +
1
2
𝛽, so that
𝜃 = 𝛼 + 𝛽.
If 𝛼 + 𝛽 > 180°, then 𝜃= ( 𝛼 + 𝛽) − 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 (<ABC) = 40˚ and m (<BAD) = 120˚, determine:
A. m (<DAB),m(BCA), m(<ECA)
B. The magnitude of angles fom
𝐴𝐵
↔ to
𝐵𝐶
↔ , from
𝐴𝐶
↔ to
𝐵𝐶
↔ , and from
𝐴𝐵
↔ to
𝐴𝐶
↔
3. Suppose point A and P are different, find m(<PAP') if P’
is the image P by the
following transformations:
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

55
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 (<PAP”
) = 118˚ what is the angle magnitude between s and t.
6. Given points A, B and B 'such that RA, α (B) = B'. Construct two lines s and t such that
MsMt = RA, α
7. Given 0 is the center point of the orthogonal coordinate system and A=(1,0). Find the
coordinates of the following points:
a. 𝑅0,90( 𝐴)
b. 𝑅0,45(𝐴)
c. 𝑅0,120( 𝐴)
d. 𝑅0,−135( 𝐴)
8. Writen the equation of lines s and t such that 𝑀𝑠 𝑀𝑡 equals the following rotation, if
A=(1,3) and 0 is the center point.
a. 𝑅 𝑂,90
b. 𝑅 𝑂,−180
c. 𝑅 𝑂,120
d. 𝑅 𝐴,90
e. 𝑅 𝐴,−90
9. If A is the center point of an orthogonal coordinate system and 𝑠 = {(𝑥, 𝑦)|𝑦 = 2𝑥 −
3} find the equation of s’
= 𝑅 𝐴,90( 𝑆)
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’=𝑅 𝐴,90( 𝐼)

56
REFERENCES
Remsing, Claudia. 2006. Transformation Geometry. Rhodes University

Diktat Geometri Transformasi (English)

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