Tolong bantu jawab! Persamaan peta garis 3x 4y =12 karena refleksi terhadap garis y - x=0 dilanjutkan oleh transformasi yang bersesuaian dengan matriks (-3 5 *

jawab

garis 
3x + 4y = 12
T1 = Reflesi y = x --> T1 [0...1||1...0] 
T2 = Trans Matriks [-3..5||-1..1]

(x'.y') = T2 o T1 (x,y)

[tex] \left[\begin{array}{ccc}x'\\y'\end{array}\right] \,= \left[\begin{array}{ccc}-3&5\\-1&1\end{array}\right] \left[\begin{array}{ccc}0&1\\1&0\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] [/tex]

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

[tex]\left[\begin{array}{ccc}x\\y\end{array}\right] \,= \left[\begin{array}{ccc}5&-3\\1&-1\end{array}\right]^{-1} \left[\begin{array}{ccc}x'\\y'\end{array}\right][/tex]

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

x = -1/2 (x' + 3y')
y = -1/2 (-x' + 5y')
substitusi ke  3x + 4y = 12

3{ -1/2 (x' + 3y') } + 4 { -1/2 (-x' + 5y')} = 12 ...kalikan (-2)
3(x + 3y) + 4(-x + 5y) = - 24
3x + 9y - 4x + 20y = - 24
-x + 29y = - 24
x - 29y = 24