Garis 16x-8y+5=0 di transformasi oleh matriks (3 -1 2 2) dilanjutkan dengan rotasi 90 derajat pusat (0,0) persamaan bayangan garis karena transformasi itu adal

Kelas            : 12
Mapel           : Matematika
Kategori      : Transformasi Geometri
Kata kunci  : transformasi, rotasi
Kode            : 12.2.5 (Kelas 12 Bab 5-Transformasi Geometri)

Garis 16x-8y+5=0 di transformasi oleh matriks (3 -1 2 2) dilanjutkan dengan rotasi 90 derajat pusat (0,0) persamaan bayangan garis karena transformasi itu adalah...

Pembahasan:
suatu garis ax+by+c di transformasi T1 dilanjutkan T2 :
[tex] \left[\begin{array}{ccc}x'\\y'\end{array}\right]=M_2\times M_1\times \left[\begin{array}{ccc}x\\y\end{array}\right] \\ \left[\begin{array}{ccc}x'\\y'\end{array}\right]= \left[\begin{array}{ccc}\cos 90^{\circ}&-\sin90^{\circ}\\\sin90^{\circ}&\cos90^{\circ}\end{array}\right] \times \left[\begin{array}{ccc}3&-1\\2&2\\\end{array}\right] \times \left[\begin{array}{ccc}x\\y\end{array}\right] \\ [/tex]
[tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right]= \left[\begin{array}{ccc}0&-1\\1&0\\\end{array}\right] \times \left[\begin{array}{ccc}3&-1\\2&2\\\end{array}\right] \times \left[\begin{array}{ccc}x\\y\end{array}\right] \\ \left[\begin{array}{ccc}x'\\y'\end{array}\right]= \left[\begin{array}{ccc}-2&-2\\3&-1\end{array}\right]\times \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}x\\y\end{array}\right]= \frac{1}{(-2)(-1)-(-2)(3)} \left[\begin{array}{ccc}-1&2\\-3&-2\end{array}\right] \left[\begin{array}{ccc}x'\\y'\end{array}\right] \\ \left[\begin{array}{ccc}x\\y\end{array}\right]= \frac{1}{8} \left[\begin{array}{ccc}-1&2\\-3&-2\end{array}\right] \left[\begin{array}{ccc}x'\\y'\end{array}\right] [/tex]
[tex]\left[\begin{array}{ccc}x\\y\end{array}\right]= \frac{1}{8} \left[\begin{array}{ccc}-x'+2y'\\-3x'-2y'\end{array}\right] \\ \left[\begin{array}{ccc}x\\y\end{array}\right]= \left[\begin{array}{ccc}- \frac{1}{8} x'+ \frac{1}{4} y'\\- \frac{3}{8} x'- \frac{1}{4} y'\end{array}\right] \\ 16x-8y+5=0 \\ 16(- \frac{1}{8} x'+ \frac{1}{4} y')-8( - \frac{3}{8} x'- \frac{1}{4} y')+5=0 \\ -2x'+4y'+3x'+2y'+5=0 \\ x'+6y'+5=0[/tex]

Jadi, persamaan bayangan garis karena transformasi itu adalah x+6y+5=0

Jawaban: E

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