1. Bayangan x2+y2-6x+8y-11 =0 oleh pencerminan terhadap garis y = x dilanjutkan oleh rotasi (0,90°) adalah... 2. Titik B (5,-1) dirorasikan terhadap P (2,3) sej

Transformasi Geometri

Nmr 1
x²+y²-6x+8y-11 = 0
↓
(x,y) —Ref.[y=x]—› (y,x)
...
(y,x) —Rot.[O,90°]—› (x',y') = (-x,y)
• x' = -x → x = -x'
• y' = y
maka bayangannya :
(-x')²+(y')²-6(-x')+8(y')-11 = 0
x'²+y'²+6x'+8y'-11 = 0

Nmr 2
B(5,-1) —Rot.[(2,3), 90°]— B'(x',y')
• x' = (5-2).Cos90° - (-1-3).Sin90° + 5
...x' = (3)(0) - (-4)(1) + 5
...x' = 9
• y' = (5-2).Sin90° + (-1-3).Cos90° + (-1)
...y' = (3)(1) + (-4)(0) - 1
...y' = 2
maka bayanganya : B'(9,2)

Nmr 3
A(3,-5) —Ref.[sb.X]— A'(3,5)