RANGKUMAN BAB VII
SETENGAH PUTARAN
Disusun untuk memenuhi tugas mata Kuliah Geometri Transformasi Dosen Pengampu : Dr. Iwan Junaedi, M.Pd
Disusun Oleh . Ika !adia
"#$#$%$&'(
&. Pradatanu K.).
"#$#$%$&%(
*. +u + uhan
"#$#$%$&-(
#. )lfha Gemilang
"#$#$%$*$(
. Kori Korina na Pusp Puspit itas asar arii
"#$ "#$# #$% $%$* $*( (
JURUSAN MATEMATIKA MATEMATIKA FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM UNIVERSITAS NEGERI SEMARANG 2010
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BAB VII
SETENGAH PUTARAN +etengah +etengah Putaran Putaran mengeliling mengelilingii seuah titik adalah suatu in/olusi. in/olusi. +uatu setenga setengah h putaran putaran men0erm men0ermink inkan an setiap setiap titik titik idang idang pada pada seuah seuah titik titik tertent tertentu u sehingga diseut 1uga pen0erminan pada suatu titik. Definisi +euah setengah putaran pada suatu titik
A
S A
adalah suatu padanan
2ang
didefinisikan untuk setiap titik pada idang seagai erikut : P ) = P' P ' sehingg P ≠ A maka S 1 ( P . )paila P≠ sehinggaa A titik tengah ruas garis &.
´ ' . PP' PP S A = A
Setenga !"ta#an a$a%a s"at" t#ansf'asi 3ukti: )kan diuktikan S A 3i1ektif. 4ntuk memuktikan S A 3i1ektif maka harus diuktikan terleih dahulu
+ur1ektif dan In1ektif. "( )kan )kan diukt diuktika ikan n S A +ur1ektif S A 4ntuk menun1ukkan
+ur1ektif,
akan
S A
ditu itun1uk 1ukkan
'
P )= P ' ∃ P ∈ V ∋ S A ( P )mil searang P' ∈ V P' ∈ V ∋ P ' =S A ( P )
jika P= A ,maka S A ( A A )= A ' = A ' ' Jadi, ∀ P ∈ V ∃ P = P =S A ( P ) P ≠ A maka Jika P≠ aka ) men1 en1adi
sum umu ruas garis
'
S A ( P P )= P ' Jadi, S A +ur1ektif "&( )kan )kan diukt diuktika ikan n S A In1ektif Missal B 1 ≠ B 2 Kasus I
B 1=B 2= A 4ntuk B 1= A 4ntuk B 2= A
S A ( B1 )= B1= B1 ' 555555..6( maka S A ( B2 )= B2= B2 ' 5555555&6( Dari 6( dan &6( maka diperoleh S A ( B1 ) ≠ S A ( B2 )
Kasus II
B ≠B ≠ A
maka
, era erart rtii
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S A ( B1 )= B1 dan '
+ehingga
S A ( B2 )= B2 '
S A ( B1 )= S A ( B2 ) S A ( B1 )= S A ( B2 )
)ndaikan Karena
B 1=S A ( B1 )= S A ( B 2) = B2 ' '
Maka
' +ehingga diperoleh B 1=B 2 ' dan ᒐ1= B2 Menurut teorama, 7Melalui dua titik han2a dapat diuat satu garis8 Ini kontradiksi dengan pern2ataan ahwa B 1 ≠ B 2 Pengandaian B 1 ≠ B 2 makaS A ( B1 )= S A ( B2 ) harus diatalkan.
Jadi, S A ( B1 ) ≠ S A ( B2 ) Jadi S A In1ektif Dari "( dan "&( maka diperoleh S A +ur1ektif dan S A In1ektif Karena S A +ur1ektif dan S A In1ektif, maka S A 3i1ektif Karena S A 3i1ektif, maka S A adalah suatu transformasi. Jadi, terukti ahwa suatu setengah putaran adalah transformasi.
Tee'a ()1 An$a An$ai* i*an an
A
+e#!&t&ngan $i
se+" se+"a a titi titi*, *,
A .
Ma*a
g $a $an S A = M g M h
h
$"a ga#is ga#is tega* tega* %"#"s %"#"s -ang -ang
h
dua dua gari gariss tega tegak k luru luruss 2ang ang
)
B"*ti .
A
Dike Diketa tahu huii
erpotongan di
se seuah uah titi titik, k,
A
P ≠ A a( Kasus I : P≠ g ⊥h Karen arenaa dengan
g
g
da dan
. maka maka dapat dapat dient dientuk uk seuah seuah sistem sistem sumu sumu ortho orthogon gonal al
seagai sumu 9 dan
h seagai sumu .
A
seagai titik
asal. )mil titik
P ∈ V
Perhatikan Gamar %.&
P' (− x , y g
P";,2(
A '
X
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Ditun1ukkan ahwa untuk setiap
S A ( P )= P ' ' maka
( 0,0 )=
(
x1 + x y 1+ y ,
Diperoleh )rtin2a
2
2
erlaku
S A ( P P )= M g M h ( P P )
S A ( P P )= P' ' ( x x 1 , y 1 )
x , y ) ≠ A dan )ndaikan P ( x Karena
P
A
titik tengah
PP' PP '
sehingga
)
x 1+ x = 0 ⟺ x 1=− x
dan
〱1+ y =0 ⟺ y 1=− y
〱 A ( P ) =( − x , − y ) 555555555555555555"(
Komposisi pen0erminan
[
M g M h ( P )= M g M M h ( P ) < <
]
M g (− x , y ) (− x , − y )
)rtin2a M g M h ( P )=(− x , − y )
55555555555555555
"&( Dari persamaan "( dan "&( diperoleh Jadi,
¿ A ( P P )= M g M h ( P P ) .
S A = M g M h
( Kasus II : P= A Menurut Definisi,
S A ( A )= A
55555555555555555
"6(
M g M h ( A ) = M g ( A A )= A
555555555555555555."&6(
Dari persamaan "6( dan "&6( diperoleh Jadi,
S A ( A ) = M g M h ( A ) .
S A = M g M h .
Tee'a ()2 Ji*a g $an h $"a ga#is -ang tega* %"#"s 'a*a ' a*a M g M h= M h M g B"*ti P ≠ A a( Kasus I : P≠ P ≠ A , maka M g M h( P )= S A ( P ) . Karena P≠ M h M g ( P P )= M h ( M M g ( P P ) ) =㜵h ( ( x, x , − y ) )=(− D ,− y ) = 〰 A ( P ) .
P ) diperoleh M g M h( P )= S A ( P )= M h M g ( P Jadi, M g M h= M h M g
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+ehingga diperoleh M g M h ( A ) = M h M g ( A ) . Jadi, M g M h= M h M g .
Tee'a ()/ Ji*a S A setenga !"ta#an, 'a*a S A= S A ) B"*ti )ndaik )ndaikan an g dan dan h dua garis garis 2ang tegak tegak lurus maka −1
M g M h=S A
A titik potong antara g dan h . ( M g M h)−1= M −1h M −1g =S−1 A . −1 −1 −1 Karena M h = M h dan M g= M g maka M h M g =S A . Karena g ⊥ h , maka menurut teorema %.&, M g M h= M h M g . +edangkan menurut teorema %., S A =てg M h . −1 +ehingga diperoleh S A= M h M g= M g M h =S A . −1 Jadi, S A= S A . dengan
Tee'a ()
x , y ) 'a*a S A ( P )=( 2 a− x , 2 b − y ) ) Ji*a A =( a , b ) $an P=( x, B"*ti P ≠ A a( Kasus I : P≠ S A ( P )= P maka Misa Misalk lkan an P =( {x} rsub {1} , {y} rsub {1} ) da dan A
titik tengah
P P sehingga diperoleh y 1+ y
(( ) ,( ))
(a , b) = Maka
x 1 + x 2
y 1 + y 2
x 1 + x 2
x 1 + x 2
2
=a dan
y 1 + y 2
=b sehingga diperoleh
=a ⟺ x1 + x =2 a ⟺ x 1=2 a− x =b ⟺ y 1 + y =2 b ⟺ y 1=2 b − y
55555555555.."6( 555555555555"&6(
x 1 , y 1 )=( 2 a − x ) , ( 2 b − y ) Dari persamaan "6( dan "&6( maka ( x P )=( x x 1 , y1 ) =( 2 a− x ) , (2 b − y ) Karena S A ( P )= P , maka S A ( P Jadi, S A ( P )=( 2 a− x , 2 b − y ) . ( Kasus II : P= A Karena P= A , maka ( x , y )= ( a ,b ) artin2a a = x dan b = y . ⍞ P )= S A ( A )= A =(a , b ) A ( P
( a , b ) =( ( 2 a−a ) , ( 2 b −b ) ) ¿ ( ( 2 a− x ) , ( 2 b − y ) )
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()2 Lan"tan Setenga P"ta#an
Kita ingat kemali tentang refleksi atau pen0erminan. Definisi refleksi atau pen0erminan ialah .
M g ( A)
= A, A ∈ g
&.
M g ( P )
= P = , 2ang ersifat g adalah sumu ruas garis PP =
Jelas ahwa ∀ A ∈ g 2ang di0erminkan terhadap garis g maka ) erimpit dengan petan2a. Titik Titik 2ang demikian dinamakan titik tetap "in/ariant( refleksi.
Definisi A $ina'a*an titi* teta! in3a#iant4 t#ansf'asi T a!a+i%a +e#%a*" TA4 5 A
Dari definisi terseut, kita dapat memperoleh fakta ahwa seuah refleksi garis g memiliki tak hingga an2akn2a titik tetap 2aitu semua titik pada sumu refleksi g itu sendiri. +edangkan pada seuah setengah putaran di P "+ p(, maka satu>satun2a
∉ P dan P titik titik /arian adalah P, sea + p"P( < P dan + p"9( < 9? dengan X ∉ tengah ruas garis XX = . Definisi Se+"a t#ansf'asi T -ang +e#sifat +a6a se+"a ga#is !etan-a "ga ga#is $ina'a*an *&%ineasi
Karena setiap isometri0 adalah suatu kolineasi maka refleksi dan setengah putaran adalah suatu kolineasi. Diantara kolineasi terseut ada 2ang diseut dilatasi
Definisi S"at" *&%ineasi $ina'a*an s"at" $i%atasi sifat
∆ ( g )/ ¿ g
Tee'a Te e'a ()7
)
∆ i*a "nt"* setia! ga#is g +e#%a*"
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: +) seuah garis g, A ∉ g
Diketahui
3uktikan ahwa
S A ( g )/ ¿ g
3ukti : Misal
P ∈ g,danQ ∈ g
kare karena na
Q∈ g
P ∈ g
maka maka ) titik titik tengah tengah
maka ) titik tengah @
QQ
'
P P'
dengan
P' =S A ( P P )
denga dengan n
karena
Q' = S A ( Q )
P )
∆ APQ APQ ' dan dan ∆ AQP AQP '
Perhatikan 4ntuk
memuktikan
∆ APQ APQ ' dan dan ∆ AQP AQP ' m ( ¿ PA PA Q
'
ahwa
g' g
maka maka
adalah kongruen.
)= m( ¿ QA P ' )
"sudut ertolak elakang(
PA PA= AP '
" karena ) titik tengah
P P'
(
QA = AQ
" karena ) titik tengah
QQ
'
(
Menurut definisi kekongruenan "+ +d +( sehingga Karena Karena
∆ APQ APQ ' ≅∆ AQP'
∆ APQ' ∆AQP ' maka PQ P Q' = Q P' PQ P Q' = Q P' maka g' g ≅
haru haruss
ditu ditun1 n1uk ukka kan n
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A ∉ g , A ∉ h Ditan2a
: tentukan semua X ∈ g , Y ∈ h ∋ A titik ten gah 9:
Jawa
: )mil P ∈ g Jika P = = S A ( P ) maka g = = S A ( g ) melalui P’ melalui P’ dan P dan PA A=AP’, g’//g Jika g? memotong h di Tarik YA memotong g di 9 Maka 9 dan pasangan titik 2ang di0ari Ilustrasi :
h P?
g?
)
P
9
g
Dari Dari 0ont 0ontoh oh di atas, atas, ukt uktik ikan an ahw ahwaa 9 dan dan satu satu>sa >satu tun2 n2aa pasan pasanga gan n 2ang 2ang memenuhi memenuhi pers2aratan, pers2aratan, dan 1ika tidak menggunakan menggunakan g = = S A ( g ) tapi tapi h= = = S A ( h ) apakah akan memperoleh pasangan lain lalu 1elaskan hal terseut
Dipun2ai
: garis g dan h tidak se1a1ar
A ∉ g , A ∉ h , Ditan2a
: )d 9 dan satu>satun2a pasangan 2ang memenuhi pers2aratan.
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S A ( h ) =h' h
karena
, maka
S A ( X X ) =# ∈ h
Karena titik potong dari dua garis atau leih akan han2a ada satu titik potong, Maka
X
dan
#
satu>satun2a pasangan .
X ∈ h ' , X ∈ g , X ∈ X# , dan dan# # ∈ h , # ∈ g' , # ∈ X# X dan # satu>satun2a pasangan.
sehingga 1adi,
Dipun2ai
: garis g dan h tidak se1a1ar
A ∉ g , A ∉ h , h= = = S A ( h ) Ditan2a Ditan2a
: )pakah ada pasangan pasangan lain 2ang memenuhi memenuhi pers2aratan pers2aratan selain 9 dan .
3ukti
:
Tee'a ()8 Hasi% *a%i $"a setenga !"ta#an $engan !"sat -ang +e#+e$a, ti$a* 'e'i%i*i titi* teta!
3ukti
:
Misal A, B ∈ V , A ≠ B )kan diuktikan S A S B tidak memiliki titik tetap Misal g < AB h
AB di ), k
AB di 3
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=
M g M h M g M k
[ M M M ] M = [ M M M ] M = [ M ( M M ) ] M =
g
h
g
k
h
g
g
k
h
g
g
k
= ( M h I ) M k = M h M k
)kan ditun1ukkan S A S B tidak memiliki titik tetap Misal 9 titik /arian S A S B
( M h M k ) ( X ) Jadi S A S B "9( < 9 sehingga
= X
Jadi
M h [ ( M h M k ) ]( X ) = M h " X ( ... ()
[ ( M h M h ) M k ]( X ) = M h " X (
...( &)
Dari "( dan "&( diperoleh
M h ( X ) Misal
= IM k ( X ) ⇔ M h ( X ) = M k ( X )
M k ( X )
"i(
= X
Kasus " X ≠ X ( Misal X ≠ X
→h≠
k
Karena h dan k adalah sumu ruas garis 99 dan ruas garis han2a memiliki satu sumu maka h
Kasus & " X = X ( Misal X = X
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Ilustrasi teorema %.' k
h g )
3
Tee'a ()( Ji*a Ji*a A ≠ B a$a%a $"a titi* 'a*a an-a a$a sat" setenga !"ta#an -ang 'e'eta*an A !a$a B
3ukti : Dipun2ai A ≠ B )kan diuktikan S T ( A)
=
B dengan T titik tengah ruas garis AB
Misal ada dua setengah putaran + D dan +E sehingga S D ( A) Jadi S D ( A)
= B dan
+E ( A) = B
= +E ( A)
− − Maka S D [ S D ( A) ] = S D [ +E ( A) ]
Karena +>D<+D maka A = S D [ +E ( A) ] Jadi 1ika D ≠ E , maka erarti ahwa ) adalah titik tetap dari S D S E Cal ini tidak mungkin ada leih dari satu setengah putaran 2ang memetakan ) pada 3. +atu>satun2a setengah putaran adalah + T")( < 3 dengan T titik tengah ruas garis AB
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3ukti : "( "( Jela Jelass +P"g( < g? suatu garis. Misal A ∈ g , B ∈ g Maka A ∈ g = , B ∈ g = dan P) < P)?, P3 < P3? P) < P)? P)?, P3 < P3?, P3?, dan dan m( ∠ APB )
Kare Karena na
=
m( ∠ A= PB =) sehingga
PAB B ≅ ∆ PA PA= B " s sd s) ∆ PA Jelas m( ∠ B = A= P )
=
m( ∠ BAP )
Jadi g+P"g( dan +P seuah dilatasi "&( "&( Karen Karenaa
S p S p ( A)
= S p ( A=) = A , maka ∀ A ∈ g ∋ S P S P ( g ) = I ( g )
Jadi, S P S P = I . Cal ini erarti +P ersifat in/oluntorik Dari pern2ataan "( dan "&( diperoleh fakta ahwa + P seuah dilatasi ersifat in/olu in/olunto ntorik rik.. )tau )tau dengan dengan kata lain lain suatu suatu setenga setengah h putaran putaran adalah adalah suatu suatu dilatasi 2ang ersifat in/olutorik. Ilustrasi
: 3
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( ⇒) )mil A ∈ T ( H ) Jadi ∃ X ∈ H ∋ A = T ( X ) −
maka T ( A)
= T −[T ( X ) ] = (T −T )( X ) = I ( X ) = X
− Jadi, T ( A ) ∈ H
( ⇐) )mil T − ( A) ∈ H − Cal ini erarti T T ( A)
∈ T ( H ) atau ) ∈ T( C)
Aontoh : Dipun2ai
:
E = ( x, y ) x & + # y & = '
Misal ) < "#,>*( dan A < "*,( g adalah adalah sumu sumu 9 Ditan2a
: +elidiki apakah
Jawa
:
A ∈ M g S c ( E )
( M g S c ) = S c M g = S c M g Jelas −
)mil P < ";,2(
−
−
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Menu Menuru rutt teor teorem emaa %.B, %.B, A ∈ T ( H )
M g S c ( E )
↔ T − ( A) ∈ H .
−
c
Jadi haruslah
maka
= ( ' − x,& + y )
( M S ) ( P ) ∈ E ↔ ( ' − x,& + y ) ∈ ( x, y ) x −
Jadi,
,
↔ ( M g S c ) − ( P ) ∈ E . 3erdasarkan perhitungan 2ang telah dilakukan
( M g S c ) ( P ) seelumn2a, 1ika P = ( x, y ) maka g
tran transfo sform rmasi asi T adal adalah ah
E = ( x, y ) x & + # y & = '
de dengan
P ∈ M g S c ( E )
Jika Jika
&
&
+ # y = '
( ' − x) & + #( & + y ) & = '
Cal ini erarti ahwa
P ∈ M g S c ( E )
↔ P ( x, y ) ∈ { x & + # y & − & x + ' y + *' = $}
& +ehingga diperoleh fakta ahwa x
peta E oleh transformasi
M g S c
+ # y & − & x + ' y + *' = $ adalah persamaan
.
Latian S&a% a%a'an 89 . Dike Dikett : tit titik ik ), 3, 3, P tak tak seg segari ariss dan dan er ered eda. a. Fukis : S ( P P ) a.
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)
S A S B ( P P ) d.
3 P
S B S A ( P P )
S A ( P P ) ) 3 P
e.
S B S A ( P P )
S A2 ( P ) S A ( P P ) )
3 2 < ¿ S A ( P ) &. Dik Diket : garis aris g dan titik A , A ∉ g P Ditan2a : a( Fuki Fukisa san n gar garis is g1= S A ( g ) dan mengapa ( 3uktikan ahwa g' / ¿ g . Jawa :
g
seuah garisH
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4ntuk memuktikan ahwa
∆ APQ' APQ' dan dan ∆ AQP' AQP'
g' g
maka harus ditun1ukkan
adalah kongruen.
"sudut ertolak elakang( m ( ¿ PA PA Q' )= m( ¿ QA P' ) PA PA= AP ' " karena ) titik tengah P P' (
Q' A = AQ
" karena ) titik tengah
Q Q'
(
Menurut defnisi kekongruenan (S Sd S) sehingga Karena Karena
*. Diket
:
∆ APQ APQ ' ∆ AQP' ≅
∆ APQ' ∆AQP ' maka PQ P Q' = Q P' PQ P Q' = Q P' maka g' g ≅
&X # &X#.
∆ AB% dan 1a1argen1ang
∆ AB% dan diluar 1a1argen1ang Ditan2a : a( Fukisan S ( ( ∆ AB% ) ( Titik J ∋ S ) ( &X# ) =&X# Jawa : a( Fukisan S ( (∆ AB% ) 3
A?
, K terletak diluar daerah
)?
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b)
M −1k ( % )=%
dan
S *= M k M m
. Dike Dikett : ) < "&, "&,*( *( Ditan2a: a. +)" A ( apaila A < "&,*( . +)" D ( apaila D < ">&,%( 0. +)" E ( apaila E< "#,>( d. +)" P ( apaila P < ";,2( Jawa: a. A < "&,*( +)" A ( < "&.& > &, &.* > *( < "&,*( . D < ">&,%( +)" D ( < "&.&>">&(, &.*>%( < "',>( 0. E< "#,>( +)" E ( < "&.&>#, &.*>">(( < "$,%(
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. K < ", >#( +3"K(
< "&.&>, &.' > ">#( < "*, '(
+D+3"K( < +D"*,'( < "&.$ > *, &.">*( > '( < ">*, >&&( 0. +D"K(
< "&.$ > , &.">*( > ">#(( < ">, >&(
+3+D"K( <+3">, >&( < "&.& > ">(, &.' > ">&(( < ", #( d. Menu Menuru rutt teor teorem emaa %.* %.* >
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P ( x x , y ) M g S % ( P P ) = M g ( 2. (−4 )− x , 2.3− y ) ¿ M g (−8 − x , 6 − y ) ¿ ( y −6, x +8 ) M −1 c. ( ¿ ¿ g S % ) ( P P )=( S % −1 M g−1 ) ( P ) b.
¿
Berdasarkan
teorema
7.3
dan
6.3
diperoleh
S A−1= S A dan M g− 1= M g , sehingga diperoleh M −1 ( ¿ ¿ g S % ) P P ) ( P )=( S % −1 M g−1 ) ( P ¿ ¿ ( S % M g ) ( P P )
¿ S % M - ( ᒐ , # ) )
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Misal
A=(a , b) B =( + , d )
D=( x x , y ) D )= S B ( D D ) Karena S A ( D ( 2 a− x , 2 b − y )=( 2 + − x , 2 d − y ) maka 2 a− x = 2 + − x diperoleh ⇔ 2 a =2 + ⇔ a =+ 2 b− y =2 d − y dan ⟺ 2 b =2 d ⟺ b= d Karena a =+ dan b =d Maka ( a , b ) = ( + , 敡 ) sehingga A = B Jadi dapat ditarik suatu akiat 2aitu A = B / )= / x , y ) ⟹ Misal A ( a , b ) , / ( x, c. S A ( / S A ( / / )=( 2 a − x, x , 2 b− y )
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= S B ( &a − x,&b − y ) = ( &c − ( &a − x ) ,&d − ( &b − y ) ) = ( &c − &a + x,&d − &b + y ) S B S A ( P ) = ( − &a + &c + x,−&b + &d + y )
...( &)
Dari "( dan "&( diperoleh fakta ahwa
( &a − &c + x,&b − &d + y ) ≠ ( − &a + &c + x,−&b + &d + y ) S A S B
≠ S B S A
Jadi, ∀ A ≠ B, S A S B ( Dipun2ai
=
S B S A merupakan pern2ataan 2ang salah
: setiap setengah putaran adalah suatu isometri0 langsung
Ditan2a n2a
: se selidiki ap apakah pe pern2 rn2ata ataan te terse seut e enar
Jawa
:
Menu Menuru rutt
defi defini nisi si
suat suatu u
tran transf sfor orma masi si
isom isomet etri ri00
lang langsu sung ng
apa apail ilaa
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A B = =
( ( & x& − x ) − ( & x − x& ) ) & + ( ( & y& − y ) − ( & y − y& ) ) & ( & x& − x − & x + x& ) & + ( & y& − y − & y + y& ) &
=
( * x& − * x ) & + ( * y& − * y ) &
=
B( x& − x ) + B( y& − y )
=*
&
( x& − x ) & + ( y& − y ) &
&
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g dan h ada di A (−1,0 ) A titik potong g dan h
titik potong B ( 3,4 ) ∈ g
Sehingga
A dan B
Persamaaan
B (3,4 )
garis
∈g
g me melalui
A (−1,0 )
dan
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f.
"Mg+3(>
+eles elesai aian an :
#. Diketah Diketahui:t ui:tiga iga titik titik ), 3, 3, A 3uktikan: 3ukti:
( S A SB )−1= S B S A
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Pen&elesaian( a. iket ( A!"#")# $ $!-%#1) itan&a ( tentukanlah tentukanlah K sehinga sehinga Ja*ab ( =( x , y ) +isal ( =( ⇔
(
S A S B ( ( ( ) =( 6,2 ) S A S B ( x x , y ) =( 6,2 ) )
S A S B ( ( ( ) =(6,2 )
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y =2 〰 + 13 Jadi, g' ={( x, x , y )∨ y = 2 x + 13 } . Kasus I )mil titik 2 = A 2 (−1,4 )∈ h S A ( h ) =h' , 2 ∈ h,danS A ( 2 2 )= 2 ' Maka 2 ' ∈ h ' S A ( 2 2 )=( 2. (−1 )−(−1 ) , 2.4 − 4 )
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