Problem 3: Triangle ABC with A(1,2), B(3,1), C(2,4) is rotated 90° counter-clockwise about the origin. Find A’B’C’.
Solution:
Rule for R(O, 90° CCW): (x,y) → (-y, x).
A’ = (-2, 1)
B’ = (-1, 3)
C’ = (-4, 2)
Student Difficulty: Mixing up clockwise vs. counter-clockwise rules.
Remediation: Mnemonic: “90 CCW: y becomes negative and jumps to x.”
| Transformation | Notation | Formula (Initial point ( P(x,y) ) ) | | :--- | :--- | :--- | | Translation (Pergeseran) | ( T = \beginpmatrix a \ b \endpmatrix ) | ( P'(x+a, y+b) ) | | Reflection (Pencerminan) | ( M_axis ) | Varies (see table below) | | Rotation (Perputaran) | ( R[O, \theta] ) | ( P'(x\cos\theta - y\sin\theta, x\sin\theta + y\cos\theta) ) | | Dilation (Perkalian) | ( [O, k] ) or ( [P, k] ) | ( P'(kx, ky) ) (center O) |
Soal 5: Titik ( M(3, -7) ) diputar sejauh 90° berlawanan arah jarum jam dengan pusat ( O(0,0) ). Koordinat bayangannya adalah...
Pembahasan: Gunakan rumus rotasi 90°: ( (x, y) \to (-y, x) ) [ M(3, -7) \to M'( -(-7), 3) = (7, 3) ]
Soal 6: Jika titik ( N(2, 5) ) dirotasi 180° menghasilkan ( N' ), lalu direfleksikan terhadap sumbu X, tentukan koordinat akhirnya. Soal Transformasi Geometri Kelas 9
Pembahasan: Rotasi 180°: ( (2,5) \to (-2, -5) ) Refleksi sumbu X: ( (-2, -5) \to (-2, 5) ) Jadi koordinat akhir ( (-2, 5) ).
Rotasi adalah memutar titik/bangun terhadap pusat dan searah jarum jam atau berlawanan arah jarum jam (B.A.J).
Rumus Umum (Pusat $O(0,0)$):
Contoh Soal: Tentukan bayangan titik $C(5, 2)$ jika dirotasikan sejauh $90^\circ$ berlawanan arah jarum jam dengan pusat $O(0,0)$.
Pembahasan:
Bayangan titik B setelah translasi ( \binom2-3 ) adalah B'(5, -1). Tentukan B. Problem 3: Triangle ABC with A(1,2), B(3,1), C(2,4)
Soal 3: Bayangan titik ( K(-5, 3) ) jika dicerminkan terhadap garis ( y = x ) adalah...
Pembahasan: Gunakan rumus refleksi ( y = x ): [ (x, y) \to (y, x) ] Maka: [ K(-5, 3) \to K'(3, -5) ]
Soal 4: Tentukan bayangan garis ( y = 2x + 4 ) jika dicerminkan terhadap sumbu Y.
Pembahasan: Refleksi sumbu Y: ( x' = -x ) atau ( x = -x' ), dan ( y' = y ). Substitusi ke persamaan garis: [ y = 2(-x') + 4 ] [ y = -2x' + 4 ] Jadi bayangan garis adalah ( y = -2x + 4 ).
Dilatasi adalah transformasi yang mengubah ukuran objek (membesar atau mengecil) tetapi tidak mengubah bentuk. Dilatasi dengan pusat ( O(0,0) ) dan faktor skala ( k ):
[ A(x, y) \to A'(k \cdot x, k \cdot y) ] Soal 5: Titik ( M(3, -7) ) diputar
In a quiet classroom in Yogyakarta, nine students of Class 9B were staring at a whiteboard filled with coordinate grids. Their teacher, Ibu Dewi, had just written: “ULANGAN HARIAN: TRANSLASI, REFLEKSI, ROTASI, DILATASI.”
Among them sat Bimo, who loved history but found math as confusing as a tangled thread. He looked at the sample problem:
Titik A(3,4) ditranslasikan oleh T(2,-1). Tentukan koordinat A’.
“Just move it,” he mumbled. “Two steps right, one step down. A’(5,3). That’s easy. But why does this matter in real life?”
Ibu Dewi must have read his mind. She smiled and said, “Class, your real test isn’t on paper today. It’s in the school’s old library. Someone has hidden the key to the ‘Lumbini Chest’—a box full of ancient Javanese relics. To find it, you must solve four transformation problems. Work as a team.”
The class buzzed with excitement. Bimo’s heart raced. A treasure hunt?