The document summarizes different types of geometry transformations: 1. Translations move every point on a shape by a distance and direction without changing the shape. 2. Reflections produce mirror images of shapes across lines or axes. 3. Rotations turn shapes around a fixed point by a certain angle. 4. Dilations enlarge or reduce the size of shapes around a fixed point but do not alter their form. Composing multiple transformations means applying one transformation to the output of another. Matrix representations can describe the combined effect of transformations.

1. GEOMETRY TRANSFORMATION
1.​ ​TRANSLATION
In other words, translation is a transformation that moves each point on the plane with a certain
distance and direction.
Formula
example.
Given the triangle OAB with the coordinates of the points O (0,0), A (3,0) and B (3,5).
Determine the image coordinates of the OAB triangle if translated by T = (1, 3).
Answer ;
Example.
The image point of circle x^2 + y^2 = 25 by translation T= (-1, 3) is …
Answer ;
Because of translation T = (-1, 3) then ,
x’ = x – 1 →​ x = x’ + 1.​….(1)
y’ = y + 3 →​ y = y’ – 3​…..(2)
(1) and (2) substitute to x^2 + y^2 = 25
obtained

2. (x’ + 1)^2 + (y’ – 3)^2 = 25;
So the image point is:
(x + 1)^2 + (y – 3)^2 = 25
Try!
1. Given the points A (-3,2), B (2, -5), and C (5,4). Find the image points A, B, C if translated by
T = (-2, 4)
-> Answer :
2. Given the equation for the line x - 2y + 4 = 0. Determine the image of the line if translated by
T = (2, 3)
-> Answer :

3. 2. REFLECTION
Reflection is a transformation that moves any point on a shape to a point that is symmetrical to
the original point on the axis of the reflection.
In plane geometry, in reflection is used
a. X axis
b. Y axis
c. x = m
d. y = n
e. y = x
f. y = -x
g. Center point O (0,0)
Explains :
a. Reflection across the x axis
Based on this figure, if the image of the point P (x, y) is P '(x', y ') then P' (x ', y') = P '(x, -y) so
that the matrix form can be written as follows :
x '= x
y '= -y

4. So is the matrix reflection about the x-axis.
Example.
1. Given the triangle ABC with the coordinates of points A (2,0), B (0, -5) and C (-3.1).
Determine the image coordinates of the ABC triangle if reflection about the x-axis.
2. The image coordinate line 3x - 2y + 5 = 0 by reflection about the x axis is?
Answers.
1. reflection about the x-axis.
P(x,y) -> P’(x, -y)
A(2,0) -> A’(2,0)
B(0,-5) -> B’ (0,5)
C(-3,1) -> C’ (-3,-1)
2. by reflection about the X axis
then: x'= x -> x = x'
y'= -y -> y = -y'
x = x' and y = -y' substituted for the line 3x - 2y + 5 = 0, we get:
3x'- 2 (-y') + 5 = 0
3x' + 2y' + 5 = 0
So the image coordinate is 3x + 2y + 5 = 0
​b. Reflection across the y axis
Based on the figure, if the image of the point P (x, y) is P '(x', y ') then P' (x ', y') = P '(- x, y), so
that in matrix form it can be written as the following:
x '= -x
y '= y

5. So is the matrix reflection about the y-axis.
Example.
1. Find the image of coordinates for y = x^2 - x curve by the reflection on the Y axis.
Answer.
1. by the reflection on the Y axis then: x '= -x → x = -x' ; y '= y → y = y ’
x = -x 'and y = y' are substituted for y = x^2 - x
obtained: y '= (-x') 2 - (-x ')
y '= (x') 2 + x'
So the image is y = x^2 + x
C. ​Reflection across the line x = m
Based on the figure, if the image of the point P (x, y) is P'(x', y') then​ P' (x', y') = P'(2m-x, y).
Example.
Find the image of the curve y^2 = x - 5 by the reflection on the line x = 3.
Answer:
by reflection on the line x = 3
then: x '= 2m - x → x = 2.3 - x' = 6 –x '
y '= y → y = y ’
x = 6 - x 'and y = y' are substituted for y^2 = x - 5
obtained: (y ') 2 = (6 - x') - 5
(y ') 2 = 1 - x'
So the image is y^2 = 1 - x

6. D. ​Reflection across the line y = n
Based on the picture above, if the image of the point P (x, y) is P '(x', y ') then P' (x ', y') =
P '(x, 2n-y).
Example.
Find the image of the curve x^2 + y^2 = 4 by the reflection on the line y = -3.
Answer:
by reflection on the line y = - 3 then:
x '= x
y '= 2n - y
reflection of the line y = - 3
then: x '= x -> x = x'
y '= 2n - y
y '= 2 (-3) - y
y '= - 6 - y -> y = -y' - 6
substituted for x^2 + y^2 = 4
(x ')^2 + (-y' - 6)^2 = 4
(x ')^2 + ((- y')^2 + 12y'+ 36) - 4 = 0
So the image:
x^2 + y^2 + 12y + 32 = 0
e. ​ ​Reflection on the line y = x

7. Based on the picture above, if the image of P (x, y) is P '(x', y ') then P' (x', y') = P '(y, x), so the
matrix form can be written as follows:
x '= y
y’ = x
So is the reflection matrix with respect to the line y = x.
Ex.
The line image 2x - y + 5 = 0 which is reflected on the line y = x is….
Answer:
The reflection transformation matrix with respect to y = x is x '= y
y’ = x
So that x '= y and y' = x
substituted for 2x - y + 5 = 0
obtained: 2y '- x' + 5 = 0
-x '+ 2y' + 5 = 0
-x '+ 2y' + 5 = 0
multiplied (-1) → x '- 2y' - 5 = 0
So the image is
x - 2y - 5 = 0
F. ​Reflection on the line y = -x

8. Based on the picture above, if the image of P (x, y) is P '(x', y ') then P' (x ', y') = P '(- y, -x), so
that in matrix form it can be written as following:
x '= -y
y '= -x
So is the reflection matrix with respect to the line y = -x.
Ex.
The image of the circle equation x^2 + y^2 - 8y + 7 = 0 which is reflected on the line y = -x is….
Answer:
x '= -y and y' = -x or y = -x 'and x = -y'
Then substituted to
x^2 + y^2 - 8y + 7 = 0
(-y ')^ 2 + (-x)^2 - 8 (-x) + 7 = 0
(y ')^2 + (x')^2 + 8x + 7 = 0
(x ')^2 + (y')^2 + 8x + 7 = 0
So the image is
X^2 + y^2 + 8x + 7 = 0

9. 3. ​ROTATION
is a cycle. Rotation is determined by the center of rotation and the angle of rotation.
Center Rotation O (0,0)
Point P (x, y) is rotated a counterclockwise to center and the image P' (x', y') is obtained
then: x ' = x cos a - y sin a
y ' = x sin a + y cos a

10. If the angle of rotation a = ½π (the rotation is denoted by R ½ π)
then x '= - y and y' = x
in the form of a matrix:
So R ½π =

11. Example.
1. The image equation for the line x + y = 6 after being rotated at the base of the
coordinates with the angle of rotation 90 degree, is….
Answer:
R + 90 means: x '= -y → y = -x'
y '= x → x = y'
substituted for: x + y = 6
y'+ (-x') = 6
y'- x' = 6 → x '- y' = -6
So the image: ​x - y = -6
2. The line image equation 2x - y + 6 = 0 after being rotated at the base of the coordinates
with a rotation angle of -90, is ..
Answer:
R-90 means:
x '= x cos (-90) - y sin (-90)
y '= x sin (-90) + y cos (-90)
x '= 0 - y (-1) = y
y '= x (-1) + 0 = -x
or with a matrix:

12. R-90 means: x '= y → y = x'
y '= -x → x = -y'
substituted for: 2x - y + 6 = 0
2 (-y ') - x' + 6 = 0
-2y '- x' + 6 = 0
x '+ 2y' - 6 = 0
So the image: x + 2y - 6 = 0
If the angle of rotation a = π (the rotation is denoted by H)
then x '= - x and y' = -y
3. The parabola image equation y = 3x^2 - 6x + 1 after being rotated at the base of the
coordinates with a rotation angle of 180 degree, is ..............
Answer :
H : x’ = -x → x = -x’
y’ = -y → y = -y’
substituted to
: y = 3x2 – 6x + 1
-y’= 3(-x’)^2 – 6(-x’) + 1
-y’ = 3(x’)^2 + 6x + 1 (dikali -1)
The image:
y = -3x^2 – 6x - 1

13. Exercise : ​Clockwise and Counterclockwise (mathsisfun.com)
1. Line m : 3x+4y+12 = 0 reflected against y-axis. The result of the reflection of line m is …
2. Find the image y = 5x + 4 by rotation of R (O, -90).
3. Find the image of the point (5, -3) by rotation of R (P, 90) with the coordinates of the
point P (-1, 2)!
(check your answer with ​Rotations – GeoGebra​ )

14. 4. Point D (3, -4) is reflected by y = -x and continues reflected by y-axis. The image of point
D’ is …
4. DILATION
Is a transformation that changes the size (enlarges or reduces) a shape but does not change
the shape.
Dilation Center O (0,0) and scale factor k
If point P (x, y) is dilated to center O (0,0) and scale factor k is obtained image P' (x', y') then
x' = kx and y'= ky and denoted by [O, k] .
Example.
The line ​2x - 3y = 6​ intersects the X axis at A and intersects the y axis at B. Due to dilation of [O,
-2], point A becomes A' and point B becomes B'.
Calculate the area of ​​triangle OA'B'
Answer:
line 2x - 3y = 6 intersects the X axis at​ A (3,0)​ intersects the Y axis at ​B (0, -2)​ due to dilation [O,
-2] then ​ A '(kx, ky) → A' (- 6,0) ​and​ ​B’(​k​x,​k​y) → B’(0,4)

15. Point A '(- 6,0), B' (0,4) and point O (0,0) form a triangle as shown:
So that the area : = ½ x OA’ x OB’
= ½ x 6 x 4 = 12

16. Dilation with Center P (a, b) and scale factor k
the image is
x '= k (x - a) + a and
y '= k (y - b) + b
denoted by [P (a, b), k]
Example. Point A (-5,13) is dilated by [P, ⅔] to give A '. If the coordinates of point P are (1, -2),
then the coordinates of point A 'are….
Answer:
[P (a, b), k]
A (x, y) A '(x', y ')
x '= k (x - a) + a
y '= k (y - b) + b
[P (1, -2),]
A (-5,13) A '(x' y ')
x '= ⅔ (-5 - 1) + 1 = -3
y '= ⅔ (13 - (-2)) + (-2) = 8
So the coordinates of poin​t A '(- 3,8)

17. Composition (Sequences) of Transformations
When two or more transformations are combined to form a new transformation, the result is
called a composition of transformations, or a sequence of transformations. In a composition,
one transformation produces an image upon which the other transformation is then performed.
If T1 is a transformation from point A (x, y) to point A'(x', y') followed by transformation T2 is the
transformation from point A' (x', y') to point A"(x" , y ”) then the two successive transformations
are called the Composition Transformation and are written T2 to T1.
Transformation composition with a matrix
If T1 is represented by a matrix and T2 is a matrix , the first two
transformations of T1 followed by T2 are written T2 o T1 = .
Example.

18. 1. The matrix corresponding to the​ dilation with the center (0,0) and a scale factor of 3​ followed
by ​a reflection on the line y = x​ is…
Answer:
M1 = 3 scale dilation matrix is
M2 = Matrix of reflection with respect to y = x is
The matrix corresponding to M1 is followed by M2 written ​M2 o M1​ =
So the matrix is .
2. The image of triangle ABC, with A (2,1), B (6,1), C (5,3) because the ​reflection on the Y axis
followed by​ rotation is…
0, π)
(
Answer:
Exercise.
1. Find the image of the line 10x - 5y + 3 = 0 by the transformation corresponding to
followed by

19. Answer :

20. 2. T1 is a transformation corresponding to a matrix and T2 is a transformation
corresponding to a matrix . The image of point A (m, n) by the
transformation of T1 followed by T2 is A '(- 9,7). Determine the value of m - 2n