Tuliskan rumus Transformasi Geometri

berikut ini

-translasi
-refleksi
-rotasi
-dilatasi

Tuliskan rumus Transformasi Geometri

Jawab:

Penjelasan dengan langkah-langkah:

– Translasi:

$bold{T}(T_x,T_y) = left[begin{array}{ccc}1&0&T_x\0&1&T_y\0&0&1end{array}right]$

– Rotasi :

$bold{R}(theta) = left[begin{array}{ccc}cos(theta)&-sin(theta)&0\sin(theta)&cos(theta)&0\0&0&1end{array}right]$

– Rotasi dengan pusat (a,b):

$bold{R}(theta,(a,b)) = bold{T}(a,b)cdotbold{R}(theta)cdotbold{T}(-a,-b)\bold{R}(theta,(a,b)) =left[begin{array}{ccc}1&0&a\0&1&b\0&0&1end{array}right]left[begin{array}{ccc}cos(theta)&-sin(theta)&0\sin(theta)&cos(theta)&0\0&0&1end{array}right] left[begin{array}{ccc}1&0&-a\0&1&-b\0&0&1end{array}right]\$

$bold{R}(theta,(a,b)) = left[begin{array}{ccc}cos(theta)&-sin(theta)&a(1-cos(theta))+bsin(theta)\sin(theta)&cos(theta)&-asin(theta)+b(1-cos(theta))\0&0&1end{array}right]$

– Refleksi

(terhadap garis y = tan(Ψ)x , tan(Ψ) = m = gradien garis refleksi) :

$bold{Rf}(tan(psi)x) = left[begin{array}{ccc}cos(psi)&-sin(psi)&0\sin(psi)&cos(psi)&0\0&0&1end{array}right]$

– Refleksi (terhadap garis y = tan(Ψ)x + c) :

$bold{Rf}(tan(psi)x+c) = bold{T}(0,c)cdotbold{R}(psi)cdotbold{T}(0,-c)\bold{Rf}(tan(psi)x+c) = left[begin{array}{ccc}1&0&0\0&1&c\0&0&1end{array}right]left[begin{array}{ccc}cos(psi)&-sin(psi)&0\sin(psi)&cos(psi)&0\0&0&1end{array}right] left[begin{array}{ccc}1&0&0\0&1&-c\0&0&1end{array}right]$

$bold{R}(theta,(a,b)) = left[begin{array}{ccc}cos(psi)&-sin(psi)&csin(psi)\sin(theta)&cos(theta)&c(1-cos(psi))\0&0&1end{array}right]$

– Dilatasi :

$bold{S}(k) = left[begin{array}{ccc}k&0&0\0&k&0\0&0&1end{array}right]$

– Dilatasi dengan pusat (a,b):

$bold{S}(k,(a,b)) = bold{T}(a,b)cdotbold{S}(k)cdotbold{T}(-a,-b)\bold{S}(k,(a,b)) =left[begin{array}{ccc}1&0&a\0&1&b\0&0&1end{array}right]left[begin{array}{ccc}k&0&0\0&k&0\0&0&1end{array}right] left[begin{array}{ccc}1&0&-a\0&1&-b\0&0&1end{array}right]$
$bold{S}(k,(a,b)) =left[begin{array}{ccc}k&0&a(1-k)\0&k&b(1-k)\0&0&1end{array}right]$