bayangan titik A( 6,1 )karena rotasi 40 derajat dilanjutkan 50 derajat terhadap pusat M( 3,2)

jawab
A(x,y)=(6,1)
α +β  = 40 + 50 = 90°
pusat M(a,b)= (3,2)

[tex] \left[\begin{array}{ccc}x'\\y'\end{array}\right]\,= \left[\begin{array}{ccc}cos\,90&-sin\,90\\sin\,90&cos\,90\end{array}\right] \left[\begin{array}{ccc}x-a\\y-b\end{array}\right]\,+\, \left[\begin{array}{ccc}a\\b\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] \left[\begin{array}{ccc}6-3\\1-2\end{array}\right]\,+\, \left[\begin{array}{ccc}3\\2\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] \left[\begin{array}{ccc}3\\-1\end{array}\right]\,+\, \left[\begin{array}{ccc}3\\2\end{array}\right] [/tex]

[tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right]\,= \left[\begin{array}{ccc}1\\3\end{array}\right]\,+\, \left[\begin{array}{ccc}3\\2\end{array}\right]\,= \left[\begin{array}{ccc}4\\5\end{array}\right] [/tex]