Page 54 - Bismillah_Neat

Geometri Transformasi

′ = 0 − +
′ 0 −

′ −2 0 − 1 1
′ = 0 −2 − (−3) + −3

′ = −2 0 − 1 + 1
′ 0 −2 + 3) −3

′ −2 + 2 1
′ = −2 − 6) + −3

′ = −2 + 2 + 1 = −2 + 3
′ −2 − 6 − 3 −2 − 9

′ = −2 + 2 + 1 = −2 + 3
′ −2 − 6 − 3 −2 − 9

Berdasarkan kesamaan dua matriks diperoleh
X’ = −2 + 3

-2x = x’ – 3

′
− 3
=
−2
′
= −2 − 9
′
−2 = + 9

′
+ 9
=
−2
′
′
−3 +9
Substitusi = = dan y = = ke persamaan garis 2 + 4 - 3 = 0
−2 −2
diperoleh :
′
′
−3 +9
2( ) + 4( ) - 3 = 0
−2 −2
-(x’ – 3) – 2(y’ + 9) – 3 = 0
-x + 3 - 2y’ – 9 - 3 = 0

-x - 2y’ – 9 = 0

Jadi, persamaan garis hasil dilatasi adalah -x – 2y – 9 = 0 Atau x + 2y + 9 = 0

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