$\begin{array}{ll}\\ 81.&\textrm{Bayangan untuk titik A(1,3) oleh rotasi }\\ &\textrm{dengan pusat}\: \: \textit{O}(0,0)\textrm{sejauh}\: \: 90^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-1,3)&&\textrm{d}.\quad (1,-3)\\ \textrm{b}.\quad (-1,-3)&\textrm{c}.\quad \color{red}(-3,1)&\textrm{e}.\quad (3,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\: \: 90^{\circ},\\ \textrm{maka}\qquad\qquad\: \: &\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga}\quad\qquad&\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -3\\ 1 \end{pmatrix} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 82.&\textrm{Suatu lingkaran dengan jari-jari 4 }\\ &\textrm{dengan pusat di O(0,0) dtranslasikan}\\ &\textrm{oleh}\: \: \textrm{T}=\begin{pmatrix} 2\\ -3 \end{pmatrix},\: \textrm{maka luas }\\ &\textrm{bayangan lingkaran tersebut adalah}\\ & ....\: \textrm{satuan luas}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \pi &&\textrm{d}.\quad 8\pi \\ \textrm{b}.\quad 2\pi &\textrm{c}.\quad 4\pi &\textrm{e}.\quad \color{red}16\pi \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran berpusat}\\ &\textrm{di O dengan}\: \: r=4.\: \textrm{Karena translasi adalah}\\ &\textrm{termasuk transformasi isometri(kongruen)}\\ &\textrm{maka jari-jari lingkaran bayangannya }\\ &\textrm{akan sama dengan bendanya. Sehingga}\\ &\textrm{ luas bayangan lingkarannya}\\ &=\pi r^{2}=\pi \times 4^{2}=\color{red}16\pi \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 83.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\textrm{Yang merupakan titik invarian (tidak berubah) }\\ &\textrm{adalah}\: ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (0,0)&&\textrm{d}.\quad (0,-1)\\ \textrm{b}.\quad \color{red}(1,-1)&\textrm{c}.\quad (1,0)&\textrm{e}.\quad (1,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}:\\ &\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{NO}&\textrm{Titik}&\begin{aligned}&\textrm{Disubstitusikan ke}\\ & \color{blue}\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases} \end{aligned}&\begin{aligned}&\textrm{Keterangan}\\ &\quad\textrm{Titik} \end{aligned}\\\hline \textrm{a}.&(0,0)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(0)-4=-4 \end{cases}&\textrm{Varian}\\\hline \textrm{b}&(1,-1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(-1)-4=-1 \end{cases}&\color{red}\textbf{Invarian}\\\hline \textrm{c}&(1,0)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(0)-4=-2 \end{cases}&\textrm{Varian}\\\hline \textrm{d}&(0,-1)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(-1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \textrm{e}&(1,1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \end{array} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 84.&\textrm{Bayangan untuk titik P(2,5) oleh rotasi }\\ &\textrm{dengan pusat}\: \textit{A}(1,3)\: \: \textrm{sejauh}\: \: 180^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (1,0)&&\textrm{d}.\quad (2,0)\\ \textrm{b}.\quad \color{red}(0,1)&\textrm{c}.\quad (0,2)&\textrm{e}.\quad (1,2) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Karena rotasi de}&\textrm{ngan pusat A sebesar}\: \: 180^{\circ},\\ \textrm{maka}\qquad\qquad\: \: \: \: &\\ R\left ( A(1,3),180^{\circ} \right )&=\begin{pmatrix} \cos 180^{\circ} & -\sin 180^{\circ}\\ \sin 180^{\circ} & \cos 180^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{sehingga bayang}&\textrm{an titik P(2,5)-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\color{blue}\begin{pmatrix} x-a\\ y-b \end{pmatrix}\color{black}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 2-1\\ 5-3 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -1\\ -2 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 0\\ 1 \end{pmatrix} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 85.&\textrm{Bayangan kurva}\: \: xy=6\: \: \textrm{oleh rotasi sebesar}\\ & \displaystyle \frac{\pi }{2}\: \: \textrm{dengan pusat}\: \: O(0,0)\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}xy=-6&&\textrm{d}.\quad x(y-x)=6\\ \textrm{b}.\quad xy=6&&\textrm{e}.\quad x(x+y)=-6\\ \textrm{c}.\quad x(x-y)=6&\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\\ \displaystyle \frac{\pi }{2}=90^{\circ},\: \: \textrm{maka}&\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga bayan}&\textrm{gan semua titik yang }\\ \textrm{terletak pada k}& \textrm{urva adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =-x' \end{cases}\\ \textrm{Selanjunya}\: \: \: \textrm{unt}&\textrm{uk bayangan kurvanya }\\ \textrm{adalah}:\qquad\quad&\\ xy&=6\\ y'.(-x')&=6\\ x'y'&=-6\\ \textrm{Jadi , persamaa}&\textrm{n kurva bayangannya}\\ \textrm{adalah}\: &\: \color{red}xy=-6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 86.&\textrm{Sebuah lingkaran yang berpusat di (3,4) }\\ &\textrm{dan menyinggung sumbu-X dicerminkan}\\ &\textrm{terhadap garis}\: \: y=x\: \textrm{, maka persamaan }\\ &\textrm{akhir lingkaran yang terjadi adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x^{2}+y^{2}-8x-6y+9=0&&\\ \textrm{b}.\quad x^{2}+y^{2}+8x+6y+9=0&\\ \textrm{c}.\quad x^{2}+y^{2}+6x+8y+9=0&\\ \textrm{d}.\quad x^{2}+y^{2}-8x-6y+16=0\\ \textrm{e}.\quad x^{2}+y^{2}+8x+6y+16=0\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Refleksi l}&\textrm{ingkaran yang berpusat di (3,4) }\\ \textrm{dan men}&\textrm{yinggung sumbu-X, }\\ \textrm{dengan}\: \: r&=(y)=4,\\ \textrm{maka}\: \textrm{per}&\textrm{samaan lingkarannya adalah}:\\ (x-3)^{2}+&(y-4)^{2}=4^{2}.\: \textrm{Karena}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =x' \end{cases}\\ \textrm{selanjutn}&\textrm{ya untuk persamaan bayangan }\\ \textrm{lingkaran} &\textrm{nya adalah}:\\ &(y'-3)^{2}+(x'-4)^{2}=4^{2},\\ & \textbf{menjadi}\\ &(y-3)^{2}+(x-4)^{2}=4^{2},\quad \textrm{atau}:\\ &\color{red}x^{2}+y^{2}-8x-6y+9=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 87.&\textrm{Jika}\: \: M_{x}\: \: \textrm{adalah pencerminan terhadap sumbu-X }\\ &\textrm{dan}\: \: M_{y=x}\: \: \textrm{adalah pencerminan terhadap garis}\\ & y=x\: ,\: \textrm{maka matriks transformasi tunggal }\\ &\textrm{yang mewakili}\: \: M_{x}\circ M_{y=x}=\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{b}.\quad \color{red}\begin{pmatrix} 0 &1 \\ -1 & 0 \end{pmatrix}&&\textrm{e}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\ \textrm{c}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}\: &\textrm{bahwa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ M_{y=x} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{cases}\\ M_{x}\circ M_{y=x}&=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ &=\begin{pmatrix} 0+0 & 1+0\\ 0-1 & 0+0 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 88.&\textrm{Diketahui vektor}\: \: \vec{x}\: \: \textrm{dirotasikan terhadap titik asal}\\ & O\: \: \textrm{sebesar}\: \: \theta >0\: \: \textrm{searah jarum jam}.\\ &\textrm{Kemudian hasilnya dicerminkan terhadap garis}\: \: y=0\\ & \textrm{menghasilkan vektor}\: \: \vec{y}.\\ &\textrm{Jika}\: \: \vec{y}=A.\vec{x}\: ,\: \textrm{maka matriks}\: \: A-\textrm{nya adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}&&\\ \textrm{b}.\quad \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}&&\\ \textrm{c}.\quad \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}&&\\ \textrm{d}.\quad \color{red}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui bah}\textrm{wa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ R_{-\theta } & =\begin{pmatrix} \cos (-\theta ) & -\sin (-\theta )\\ \sin (-\theta ) & \cos (-\theta ) \end{pmatrix}=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{cases}\\ &A=M_{x}\circ R_{-\theta }=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 89.&\textrm{Titik A(1,-2) dirotasikan sejauh}\: \: 15^{\circ}\\ & \textrm{kemudian dilanjutkan}\: \: 75^{\circ}\: \: \textrm{dengan pusat }\\ &O(0,0)\: \: \textrm{maka bayangan akhir titik A adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-2,1)&&\textrm{d}.\quad \color{red}(2,1)\\ \textrm{b}.\quad (-1,2)&\textrm{c}.\quad (1,2)&\textrm{e}.\quad (-2,-1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos (\theta _{1}+\theta _{2}) & -\sin (\theta _{1}+\theta _{2})\\ \sin (\theta _{1}+\theta _{2}) & \cos (\theta _{1}+\theta _{2}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos (75^{\circ}+15^{\circ})& -\sin (75^{\circ}+15^{\circ})\\ \sin (75^{\circ}+15^{\circ}) & \cos (75^{\circ}+15^{\circ}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ}&-\sin 90^{\circ}\\ \sin 90^{\circ}&\cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 1 \end{pmatrix} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 90.&\textrm{Jika garis}\: \: 3x-2y+5=0\: \: \textrm{dicerminkan }\\ &\textrm{terhadap garis}\: \: y=-x\: \: \textrm{kemudian}\\ &\textrm{didilatasikan dengan pusat (1,-2) }\\ &\textrm{dengan faktor skala 2, maka persamaan}\\ & \textrm{bayangannya adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x-2y-10=0&&\\ \textrm{b}.\quad x+2y-10=0&\\ \textrm{c}.\quad x-6y+5=0&\\ \textrm{d}.\quad x+2y-12=0\\ \textrm{e}.\quad \color{red}2x-3y+18=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Proses}&\: \textrm{untuk refleksinya}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&-1\\ -1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ -x \end{pmatrix}\\ \textrm{proses}&\: \textrm{dilatasinya}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}\begin{pmatrix} x'-1\\ y'+2 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-2\\ 2y'+4 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-1\\ 2y'+2 \end{pmatrix}\\ &=\begin{pmatrix} 2(-y)-1\\ 2(-x)+2 \end{pmatrix}\\ &\begin{cases} x &=-\displaystyle \frac{1}{2}(y''-2) \\ y &=-\displaystyle \frac{1}{2}(x''+1) \end{cases} \end{aligned}\\ &\begin{aligned}\textrm{Sehingga persam}&\textrm{aan bayangan}\\ \textrm{garisnya adalah}:&\\ 3x&-2y+5=0\\ 3\left ( -\displaystyle \frac{1}{2}(y''-2) \right )&-2\left ( -\displaystyle \frac{1}{2}(x''+1) \right )+5=0\\ -\displaystyle \frac{3}{2}y''+3 &+(x''+1)+5=0\\ 2x&-3y+6+2+10=0\\ 2x&-3y+18=0 \end{aligned} \end{array}$.
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