Contoh Soal 1 Transformasi Geometri

$\begin{array}{ll}\\ 1.&\textrm{Suatu translasi yang memetakan titik P(9,8) }\\ &\textrm{ke titik}\: \: \textrm{P}'(14,-2)\: \: \textrm{adalah}\: ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 5\\ -10 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 6\\ 6 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 5\\ 6 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 23\\ -10 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} 5\\ 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\textrm{T}+\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{T}&=\begin{pmatrix} x'-x\\ y'-y \end{pmatrix}\\ &=\begin{pmatrix} 14-9\\ -2-8 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ -10 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ &\textrm{Maka bayangan titik M}(2,-1)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (7,10)&&\textrm{d}.\quad (1,10)\\ \textrm{b}.\quad (10,7)&\textrm{c}.\quad \color{red}(1,4)&\textrm{e}.\quad (4,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ \left.\begin{matrix} x=2\\ y=-1 \end{matrix}\right\}&\Rightarrow \begin{cases} x' & =2(2)+3(-1)=4-3=1 \\ y' & =3(2)+2(-1)=6-2=4 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Bayangan untuk titik A(1,3) oleh rotasi }\\ &\textrm{dengan pusat}\: \: \textit{O}(0,0)\textrm{sejauh}\: \: 90^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-1,3)&&\textrm{d}.\quad (1,-3)\\ \textrm{b}.\quad (-1,-3)&\textrm{c}.\quad \color{red}(-3,1)&\textrm{e}.\quad (3,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\: \: 90^{\circ},\\ \textrm{maka}\qquad\qquad\: \: &\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga}\quad\qquad&\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -3\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Suatu lingkaran dengan jari-jari 4 }\\ &\textrm{dengan pusat di O(0,0) dtranslasikan}\\ &\textrm{oleh}\: \: \textrm{T}=\begin{pmatrix} 2\\ -3 \end{pmatrix},\: \textrm{maka luas }\\ &\textrm{bayangan lingkaran tersebut adalah}\\ & ....\: \textrm{satuan luas}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \pi &&\textrm{d}.\quad 8\pi \\ \textrm{b}.\quad 2\pi &\textrm{c}.\quad 4\pi &\textrm{e}.\quad \color{red}16\pi \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran berpusat}\\ &\textrm{di O dengan}\: \: r=4.\: \textrm{Karena translasi adalah}\\ &\textrm{termasuk transformasi isometri(kongruen)}\\ &\textrm{maka jari-jari lingkaran bayangannya }\\ &\textrm{akan sama dengan bendanya. Sehingga}\\ &\textrm{ luas bayangan lingkarannya}\\ &=\pi r^{2}=\pi \times 4^{2}=\color{red}16\pi \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\textrm{Yang merupakan titik invarian (tidak berubah) }\\ &\textrm{adalah}\: ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (0,0)&&\textrm{d}.\quad (0,-1)\\ \textrm{b}.\quad \color{red}(1,-1)&\textrm{c}.\quad (1,0)&\textrm{e}.\quad (1,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}:\\ &\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{NO}&\textrm{Titik}&\begin{aligned}&\textrm{Disubstitusikan ke}\\ & \color{blue}\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases} \end{aligned}&\begin{aligned}&\textrm{Keterangan}\\ &\quad\textrm{Titik} \end{aligned}\\\hline \textrm{a}.&(0,0)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(0)-4=-4 \end{cases}&\textrm{Varian}\\\hline \textrm{b}&(1,-1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(-1)-4=-1 \end{cases}&\color{red}\textbf{Invarian}\\\hline \textrm{c}&(1,0)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(0)-4=-2 \end{cases}&\textrm{Varian}\\\hline \textrm{d}&(0,-1)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(-1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \textrm{e}&(1,1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \end{array} \end{aligned} \end{array}$