Senin, 12 Januari 2009

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Assignment 5

BAB 6

TRANSFORMASI GEOMETRI

Materi dalam bab ini:
A. Pengertian Transformasi
B. Translasi (Pergeseran)
C. Rotasi (Perputaran)
D. Refleksi (Pencerminan)
E. Dilatasi (Perkalian)
F. Komposisi dari beberapa Transformasi Geometri beserta Matriks Transformasinya

Kompetensi Dasar:
1. Menggunakan translasi dan transformasi geometri yang mempunyai matriks dalam pemecahan masalah.
2. Menentukan komposisi dari transfomasi geometri beserta matriks transformasinya.

Geometri Transformasi meliputi : Translasi, Rotasi, Refleksi, Dilatasi untuk menentukan Komposisi.

A. Pengertian Transformasi
Untuk memindahkan suatu titik atau bangun pada bidang dapat dilakukan dengan menggunakan transformasi. Transformasi geometri adalah bagian dari geometri yang membicarakan perubahan, baik perubahan letak maupun bentuk dari penyajiannya didasarkan dengan gambar dan matriks. Transformasi pada bidang ada empat macam, yaitu :
1. Translasi (Pergeseran)
2. Refleksi (Pencerminan)
3. Rotasi (Perputaran)
4. Dilatasi (Perkalian)
Translasi, refleksi dan rotasi disebut transformasi isometri artinya, bayangan(bangun hasil) dan transformasi itu kongruen dengan bangun semula. Pada dilatasi, bayangan(bangun hasil) dari transformasi sebangun dengan bangun semula, yaitu diperkecil atau diperbesar.
B. Translasi (Pergeseran)
Translasi atau pergeseran adalah perpindahan titik-titik pada bidang dengan jarak dan arah tertentu. Jarak dan arah tertentu itu diwakili oleh ruas garis berarah(vektor), misalnya AB atau dengan suatu pasangan bilangan, misalnya (a,b) yang berarti absis titik ditambah dengan a, sedangkan ordinatnya ditambah dengan b.
Suatu translasi dengan komponen T=(a,b) akan memetakan titik A(x1,y1) ke titik A’(x1+a,y1+b) yang dinotasikan dengan :
T=(a,b)=A(x1,y1)
A’(x1+a,y1+b)
Catatan:
Apabila suatu fungsi atau kurva mengalami pergeseran, fungsi persamaan atau kurva tersebut akan berubah.

Contoh 1:
Carilah bayangan (peta) titik A(4,3) dan B(5,-1) oleh translasi T=(3,2)!
Jawab:
T=(3,2) : A(4,3), A’(4+3,3+2)=A’(7,5)
T=(3,2) : B(5,-1), B’(5+3,-1+2)=B’(8,1)

Contoh 2:
Oleh translasi T=(a,b), titik A(1,-2) dipetakan menjadi titik A’(4,3).
Tentukan T!
Jawab:
T=(a,b) : A(1,-2), A’(1+A,-2+B)=A’(4,3)
1+a = 4 -2+b = 3
a=3 b=5
Jadi , T=(3,5)

Latihan 1
1. Diketahui segitiga PQR dengan titik-titik sudut P(1,2), Q(5,0) dan R(3,6).
a. Tentukan bayangan segitiga PQR oleh translasi T=(1,-3)!
b. Gambarlah segitiga PQR dan bayangannya pada sistem koordinat cartesius!
c. Tentukan persamaan garis yang melalui titik P dan R!
d. Tentukan persamaan bayangan garis yang melalui titik P dan R!
2. Suatu translasi (a,b) mengakibatkan titik P(5,-2) menjadi titik P’(2,2b)
a. Tentukan nilai a dan b!
b. Dengan translasi (a,b), tentukan bayangan titik A(2,1) dan B(-4,2)!
3. Tentukan persamaan bayangan dari:
a. garis y=2x+3 , bila ditranslasikan oleh (3,-4)
b. garis 3x+4y=6, bila ditranslasikan oleh (3,-4)
4. Tentukan persamaan peta bila lingkaran:
a. x2+y2=16 mengalami pergeseran (-1,1)
b. x2+y2-4x+8y-5=0 mengalami pergeseran (2,-1)
5. Tentukan persamaan bayangan dari kurva di bawah ini!
a. Parabola y2=4x, jika titik puncaknya ditranslasikan oleh (2,1)
b. Elips x2/9+y2/4 =1, jika titik pusatnya ditranslasikan oleh (1,-2)

C. Rotasi (Perputaran)
Rotasi atau perputaran ditentukan oleh:
1. Pusat perputaran
2. besar sudut putar, dan
3. arah sudut putar
Suatu perputaran mempunyai arah positif bila arah perputaran itu berlawanan arah dengan arah putar jarum jam. Arah negatif terjadi bila arah perputaran itu searah dengan arah putaran jarum jam.
Titik P’(x’,y’) adalah peta dari titik P(x,y) oleh rotasi terhadap O sebesar α radian. Misalkan koordinat cartesius P(x,y) ditulis dengan koordinat kutub menjadi titik P(r,θ) maka terdapat hubungan:
x = r cos θ
y = r sin θ
Titik P(x,y) = P(r,θ) diputar sebesar α radian menghasilkan P’(x’,y’) = P’(r,θ+α) sehingga:
x’ = r cos (θ+α)
= r (cos θ cos α – sin θ sin α)
= r cos θ cos α – r sin θ sin α
= x cos α – y sin c………......(1)
y’ = r sin (θ + α)
= r ( sin θ cos α + cos θ sin α)
= r sin θ cos α + r cos θ sin α
= y cos α + x sin α
= x sin α + y cos α………….(2)

Ingat!
Dari (1) dan (2) diperoleh:
cos (θ ± α ) = cos θ = sin θ sin α
sin (θ ± α ) = sin θ cos α ± cos θ sin α

Dari (1) dan (2) diperoleh:
x’ = x cos α – y sin α
y’ = x sin α + y cos α
Jadi, diperoleh rumus:
Bila titik P(x,y) diputar dengan pusat O sebesar α radian akan menjadi titik P’(x’,y’) dimana:
(x’,y’) = (x cos α – y sin α, x sin α + y cos α)
Dengan cara yang sama akan diperoleh, jika titik P(x’,y’) diputar dengan pusat A(a,b) sebesar α radian akan menjadi titik P’(x’,y’) dengan:
( x’-a , y’-b ) = ((x-a)( cos α - sin α ) , (y-b)( sin α + cos α ))

Latihan 2:
1. Tulislah matriks-matriks yang berkaitan dengan rotasi pusat O dan sudut putar sebesar:
a.+ 30˚ d.+120˚
b.+ 45˚ e.+180˚
c.+60˚ f .-150˚
2. Tentukan bayangan titik-titik sudut segitiga ABC dengan A(1,2), B(5,-2) dan C(5,6), jika dirotasikan dengan pusat O sebesar:
a. 2/3 π radian
b.3/2 π radian
3. Tentukan bayangan titik-titik berikut, jika dirotasikan dengan pusat (-1,4) dan sudut putar sebesar 90˚!
a. (2,5) c. (6,-8)
b. (-3,4) d. (-4,-6)
4. Tentukan persamaan bayangan dari hal berikut!
a. Garis 2x+y+3=0 , jika dirotasikan dengan pusat O sebesar 90˚
b. Parabola y2=4x+3, jika dirotasikan dengan pusat O sebesar 180˚
c. Lingkaran x2+y2+2x-4y=20 , jika dirotasikan dengan pusat O sebesar -90˚
5. Tentukan persamaan peta dari persamaan-persamaan berikut jika diputar dengan pusat P (-1,2) dan sudut putar 90˚!
a. Garis y =3x-1
b. Elips x2+2y=8
c. Hiperbola 4x2-9y2=1

D. Refleksi (Pencerminan)
Suatu refleksi ditentukan oleh suatu garis tertentu sabagai sumbu pencerminan yang perlu diperhatikan pada pencerminan adalah jarak bangun mula-mula ke sumbu pencerminan sama dengan jarak bangun bayangannya ke sumbu pencerminan. Sehingga pada pencerminan akan diperoleh sebagai berikut:
1. Bangun bayangan sebangun dengan bangun mula-mula.
2. Keliling bangun bayangannya sama dengan keliling bangun mula-mula.
3. Luas bangun bayangannya sama dengan luas bangun mula-mula.
Segitiga ABC dicerminkan terhadap garis α menghasilkan segitiga A’B’C’.
Maka:
AP = A’P
BQ = B’Q
CR = C’R

Pencerminan terhadap Sumbu Koordinat, Titik dan Garis Linear
Misalkan titik P(x,y) dicerminkan terhadap sumbu X menghasilkan titik P’(x’,y’)
Maka:
x’=x=(1)(x)+(0)(y)
y’=-y=(0)(y)+(-1)(y)
(x’,y’)=(x,-y)
Jadi, titik P(x,y) : P’(x’,y’) ditentukan oleh persamaan matriks:
(x’,y’)=(x,-y)

Dengan cara yang sama seperti contoh diatas, dapat diperoleh matriks-matriks pencerminan sebagai berikut:
1. (x,-y) : Pencerminan terhadap sumbu X
2. (-x,y) : Pencerminan terhadap sumbu Y
3. (-x,-y) : Pencerminan terhadap titik asal O
4. (y,x) : Pencerminan terhadap garis y=x
5. (-y,-x) : Pencerminan terhadap garis y=-x
6. ((x cos 2α + y sin 2α),(x sin 2α – y cos 2α)) : Pencerminan terhadap garis y = x tan α

Ingat!
Pencerminan suatu titik terhadap garis y=k, x=(k=0) dan garis linear selain y=x dan y-x, tidak dapat ditulis dalam bentuk matriks.

Catatan:
Cara menghafalkan matriks transformasi diatas, cukup kita ambil 2 titik pada sumbu X dan sumbu Y, dua titik yang dimaksud adalah titik A(1,0) dan B(0,1). Selanjutnya tinggal memperhatikan pernyataan soalnya.

Contoh:
Matriks yang bersesuaian dengan pencerminan terhadap sumbu X
Titik A(1,0) : A’(-1,0)
B(0,1) : B’(0,1)

In English

Chapter 6

THE GEOMETRY TRANSFORMATION


The material in the chapter is:
G. The Definition of The Transformation
H. The Translation
I. The Rotation
J. The Reflection
K. The Dilatation
L. The Composition from the several transformation of geometry and the matrix transformation

The Basic Competence :
3. Used the translation and the transformation of geometry that had the matrix in the solution.
4. Determined the composition from the transformation of geometry and the matrix transformation.

The Geometry Transformation involve of : The Translation, The Rotation, The Reflection, The Dilatation which to determine The Composition.

C. The Definition of The Transformation
To move a point or figure on the plane could be carried out with used the transformation. The transformation of geometry was part of the geometry that discussed the transform, both the transform in the location and the form and this presentation was based with the picture and matrix. The transformation on the plane had four sorts, that is :
1. The translation
2. The reflection
3. The rotation
4. The dilatation
The translation, the reflection and the rotation were acknowledged as the isometry transformation, meaning that the shadow (figure results) of the transformation was congruent. In the dilatation, the shadow (the figure results) of the transformation was congruent by the figure originally, that is reduced or dilated.

D. The Translation
The translation or the shift was the move of the points to the plane with the distance and the certain direction. The distance and the certain direction were represented by the part of the directed line (the vector),for example AB or with a pair number, for example (a,b) that was significant absis the point was increased with a ,where as ordinat him was increased with b.
A translation with the component of T=(a,b) will map the point of A(x1,y1) to the point of A’(x1+a,y1+b) that
T=(a,b)=A(x1,y1)
A’(x1+a,y1+b)
Note:
If a function or the curve experienced the shift, the regulation function or this curve will change.

Example 1:
Look for the shadow (the map) the point of A(4,3) and B(5,-1) by the translation of T=(3,2)!
Solution:
T=(3,2) : A(4,3), A’(4+3,3+2)=A’(7,5)
T=(3,2) : B(5,-1), B’(5+3,-1+2)=B’(8,1)

Example 2:
By the translation of T=(a,b)the point of A(1,-2) was mapped to the point of A’(4,3).
Determine T!
Solution:
T=(a,b) : A(1,-2), A’(1+A,-2+B)=A’(4,3)
1+a = 4 -2+b = 3
a=3 b=5
So , T=(3,5)

Exercises 1
1. Given that triangle PQR with the vertex of P(1,2), Q(5,0) and R(3,6).
a. Determine the triangle PQR shadow by the translation of T=(1,-3)!
b. Drawing triangle PQR and his shadow in the co-ordinate system cartesius!
c. Determine the line equation that through the point P and R!
d. Determine the line shadow equation that through the point P and R!
2. A translation (a,b) resulted in the point P(5,-2) become to point P’(2,2b)
a. Determine the value of a and b!
b. With the translation (a,b) , determine the point shadow of A(2,1) and B(-4,2)!
3. Determine the shadow equation of:
a. The line y=2x+3 , if it is translation by (3,-4)
b. The line 3x+4y=6 , if it is translation by (3,-4)
4. Determine the map equation when the circle:
a. x2+y2=16 experienced the shift (-1,1)
b. x2+y2-4x+8y-5=0 experienced the shift (2,-1)
5. Determine the shadow equation from the curve below this!
a. The parabola y2=4x, if this peak point is translated by (2,1)
b. The ellipse x2/9+y2/4 =1 , if this central point is translated by (1,-2)

C. The Rotation
The rotation was determined by:
1. The relation centre
2. Measure of the angle turned , and
3. The angle direction turned
A rotation had the positive direction when the rotation direction was against whit the direction turned o’clock. The negative direction happened when the rotation direction a direction with the direction turned o’clock.
The point P’(x’,y’) was the map from the point P(x,y) by the rotation toward O as big as α radian. For example the co-ordinate cartesius the point P(x,y) was written with the pole co-ordinate to the point P(r,θ) so that was gotten by relations:
x = r cos θ
y = r sin θ
The point P(x,y) = P(r,θ) was turned as big as α radian produced P’(x’,y’) = P’(r,θ+α) so that:
x’ = r cos (θ+α)
= r (cos θ cos α – sin θ sin α)
= r cos θ cos α – r sin θ sin α
= x cos α – y sin c………......(1)
y’ = r sin (θ + α)
= r ( sin θ cos α + cos θ sin α)
= r sin θ cos α + r cos θ sin α
= y cos α + x sin α
= x sin α + y cos α………….(2)

Remember!
From (1) and (2) were received:
cos (θ ± α ) = cos θ = sin θ sin α
sin (θ ± α ) = sin θ cos α ± cos θ sin α

From (1) and (2) were received:
x’ = x cos α – y sin α
y’ = x sin α + y cos α
So, was received by the formula:
When the point P(x,y) was turned with the centre O as big as α radian will become the point P’(x’,y’) where:
(x’,y’) = (x cos α – y sin α, x sin α + y cos α)
With the same way will receive , if the point P(x’,y’) was turned with the centre A(a,b) as big as α radian will become the point P’(x’,y’) with
( x’-a , y’-b ) = ((x-a)( cos α - sin α ) , (y-b)( sin α + cos α ))

Exercises 2:
1. Write the matrix which correlated with the centre rotation O and the angle turned as big as:
a.+ 30˚ d.+120˚
b.+ 45˚ e.+180˚
c.+60˚ f .-150˚
2. Determine the angle points shadow of triangle ABC with A(1,2), B(5,-2) and C(5,6), if it is rotate with the center O as big as:
a. 2/3 π radian
b.3/2 π radian
3. Determine the point shadow below this, if it is rotated with the centre (-1,4) and the angle turned as big as 90˚!
a.(2,5) c.(6,-8)
b.(-3,4) d.(-4,-6)
4. Determine the shadow equation from this below case!
a. The line 2x+y+3=0 , if it is rotated with the centre O as big as 90˚
b. The parabola y2=4x+3 , if it is rotated with the centre O as big as 90˚
c. The circle x2+y2+2x-4y=20 , if it is rotated with the centre O as big as 90˚
5. Determine the map equation from the equation below, if it is turned with the P (-1,2) and the angle turned 90˚!
a. The line y =3x-1
b. The ellips x2+2y=8
c. The hyperbola 4x2-9y2=1

D. The Reflection
A reflection was determined by a certain line as the reflection axis which need to be paid attention on the reflection was the model distance at first to the reflection axis was the same as with the model distance of his shadow to the reflection axis so on the reflection will be received as below:
1. The shadow model is congruent with the model at first
2. The circumference his shadow model is the same as with the circumference of the model at first
3. The area of his shadow model is the same as with the area of the model at first.
The triangle ABC was reflected toward the line α produced A’B’C’ triangle.
So:
AP = A’P
BQ = B’Q
CR = C’R

The Reflection toward the co-ordinate axis , the point and the linear line .
Let the point P(x,y) is reflected toward the axis x produced the paint P’(x’,y’)
So:
x’=x=(1)(x)+(0)(y)
y’=-y=(0)(y)+(-1)(y)
(x’,y’)=(x,-y)
So, the point P(x,y) : P’(x’,y’) is determined by the matrix equation:
(x’,y’)=(x,-y)

With the same way like the example above, can be received that the reflection matrix as below:
7. (x,-y) : as the reflection toward the axis x
8. (-x,y) : as the reflection toward the axis y
9. (-x,-y) : as the reflection toward the first point O
10. (y,x) : as the reflection toward the line y=x
11. (-y,-x) : as the reflection toward the line y=-x
12. ((x cos 2α + y sin 2α),(x sin 2α – y cos 2α)) : as the reflection toward the line y = x tan α

Remember!
The reflection a point toward the line y=k, x=(k=0) and the linear line aside from y=x and y-x, can not be written in the matrix form.

Note:
The procedure to memorize the transformation matrix above, only with put two points on the axis x and axis y, two points which mean the point A(1,0) and B(0,1). Furthermore remain pay attention to the problem statement.

Example:
The matrix which according to the reflection toward the axis x.
The point A(1,0) : A’(-1,0)
B(0,1) : B’(0,1)

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