Contoh Soal 1 Transformasi Geometri

$\begin{array}{l}\\ 1.& \text{Suatu translasi yang memetakan titik P(9,8) }\\ & \text{ke titik}{\text{P}}^{\prime }(14,-2)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{c}5\\ -10\end{array}\right)& & \text{d}.\left(\begin{array}{c}6\\ 6\end{array}\right)\\ \text{b}.\left(\begin{array}{c}5\\ 6\end{array}\right)& \text{c}.\left(\begin{array}{c}23\\ -10\end{array}\right)& \text{e}.\left(\begin{array}{c}5\\ 2\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\text{T}+\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{T}& =\left(\begin{array}{c}{x}^{\prime }-x\\ y^{\prime }-y\end{array}\right)\\ & =\left(\begin{array}{c}14-9\\ -2-8\end{array}\right)\\ & =\left(\begin{array}{c}5\\ -10\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ & \text{Maka bayangan titik M}(2,-1)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(7,10)& & \text{d}.(1,10)\\ \text{b}.(10,7)& \text{c}.(1,4)& \text{e}.(4,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ \begin{array}{c}x=2\\ y=-1\end{array}\}& \Rightarrow \{\begin{array}{ll}{x}^{\prime }& =2(2)+3(-1)=4-3=1\\ y^{\prime }& =3(2)+2(-1)=6-2=4\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Bayangan untuk titik A(1,3) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{O}}(0,0)\text{sejauh}{90}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(-1,3)& & \text{d}.(1,-3)\\ \text{b}.(-1,-3)& \text{c}.(-3,1)& \text{e}.(3,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Karena rotasi d}& \text{engan pusat O sebesar}{90}^{\circ },\\ \text{maka}& \\ R(O(0,0),{90}^{\circ })& =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ \text{sehingga}& \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-3\\ 1\end{array}\right)\end{array}\end{array}$.

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

$\begin{array}{l}\\ 5.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \text{Yang merupakan titik invarian (tidak berubah) }\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(0,0)& & \text{d}.(0,-1)\\ \text{b}.(1,-1)& \text{c}.(1,0)& \text{e}.(1,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \begin{array}{|cccc|}\hline \text{NO}& \text{Titik}& \begin{array}{rl}& \text{Disubstitusikan ke}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\end{array}& \begin{array}{rl}& \text{Keterangan}\\ & \text{Titik}\end{array}\\ \text{a}.& (0,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(0)-4=-4\end{array}& \text{Varian}\\ \text{b}& (1,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(-1)-4=-1\end{array}& \mathbf{\text{Invarian}}\\ \text{c}& (1,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(0)-4=-2\end{array}& \text{Varian}\\ \text{d}& (0,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(-1)-4=-3\end{array}& \text{Varian}\\ \text{e}& (1,1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(1)-4=-3\end{array}& \text{Varian}\\ \hline\end{array}\end{array}\end{array}$