Diketahui segitiga ABC dengan titik sudut A(3,1), B(-2,1), dan C(2,3). Jika segitiga ABC dirotasi sebesar (-π/2) dengan titik pusat (1,2), diperoleh bayangan se

jawab
A(3,1)
B(-2,1)
C(2.3)

Rotasi P(a,b)= (1,2) , α = - π/2

[tex]\left[\begin{array}{ccc}x'-a\\y'-b\end{array}\right]\,=\left[\begin{array}{ccc}cos(- \pi /2)&-sin(- \pi /2)\\sin(- \pi /2)&cos(- \pi /2)\end{array}\right]\, \left[\begin{array}{ccc}x-a\\y-b\end{array}\right] [/tex]

[tex]\left[\begin{array}{ccc}x'-1\\y'-2\end{array}\right]\,=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\, \left[\begin{array}{ccc}x-1\\y-2\end{array}\right] \,[/tex]

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

x' - 1 = y - 2→ x' = y - 1
y' - 2 = -x  + 1 → y' = -x - 3

A(3,1) → A' {(1-1), (-3-3)} = (0,-6)
B(-2,1) → B{(1-1),(2-3)} = (0, -1)
C(2,3) → C'{(3-1),(-2-3)} = (2,-5)