Question

Persamaan garis y+1=3x direfleksikan terhadap garis 2x-y=0, maka persamaan bayangan garis tersebut berbentuk........

Answer 1

Dicerminkan thp 2x - y = 0 => y = 2x => tan a = 2 = 2/1 = de/sa
mi = √(2^2 + 1^2) = √5 maka sin a = de/mi = 2/√5 dan cos a = sa/mi = 1/√5
sin 2a = 2 sin a cos a = 2 (2/√5) (1/√5) = 4/5
cos 2a = (cos a)^2 - (sin a)^2 = 1/5 - 4/5 = -3/5
maka matriks transformasinya :
(cos 2a .... sin 2a) (x) = (x')
(sin 2a ... -cos 2a) (y) ...(y')

(-3/5 ... 4/5) (x) = (x')
(4/5 .... 3/5) (y) ... (y')

(x) = 1/1 (3/5 ... -4/5) (x')
(y) .........(-4/5 .. -3/5) (y')

(x) = (3/5 x' - 4/5 y')
(y) ...(-4/5 x' - 3/5y')

jadi bayangan dari y + 1 = 3x adalah
(-4/5 x' - 3/5 y') + 1 = 3(3/5 x' - 4/5 y') .... kedua ruas kali 5
=> -4x' - 3y' + 5 = 9x' - 12y'
=> -13x' + 9y' + 5 = 0
=> 13x - 9y - 5 = 0