Tolong bantu jawab! D(3,-5) sudut a= 270 derajat arah berlawanan arah jarum jam pusat P (-2,3)

jawab

(x,y)  Rotasi [ P(a,b) , α ] --> (x, y')

[tex] \left[\begin{array}{ccc}x'\\y'\end{array}\right] \,= \left[\begin{array}{ccc}cos\, 270&-sin\,270\\sin\,270&cos\,270\end{array}\right] \left[\begin{array}{ccc}x- a\\y-b\end{array}\right] \,+\, \left[\begin{array}{ccc}a\\b\end{array}\right] [/tex]

D(x,y)= (3, -5)
P(a,b) =(-2, 3)
α  = 270


[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+2\\5-3\end{array}\right] \,+\, \left[\begin{array}{ccc}-2\\3\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}5\\2\end{array}\right] \,+\, \left[\begin{array}{ccc}-2\\3\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right] \,= \left[\begin{array}{ccc}2\\-5\end{array}\right] \,+\, \left[\begin{array}{ccc}-2\\3\end{array}\right]\,= \left[\begin{array}{ccc}0\\-2\end{array}\right] [/tex]