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′
Posisi akhir lukisan stelah digeser ’ , ′
′
= +
′
= +
Koordinat A setelah digeser (A’)
′
= 10 + 80 = 90
′(90, 40)
′
= 0 + 40 = 40
Koordinat B setelah digeser (B’)
′
= 80 + 80 = 160
B’(160, 40)
′
= 0 + 40 = 40
Koordinat C setelah digeser (C’)
′
= 80 + 80 = 160
C’(160, 90)
′
= 50 + 40 = 90
Koordinat D setelah digeser (D’)
′
= 10 + 80 = 90
D’(90, 90)
′
= 50 + 40 = 90
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