$\begin{array}{ll}\\ 11.&\textrm{Titik A(1,-2) dirotasikan sejauh}\: \: 15^{\circ}\\ & \textrm{kemudian dilanjutkan}\: \: 75^{\circ}\: \: \textrm{dengan pusat }\\ &O(0,0)\: \: \textrm{maka bayangan akhir titik A adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-2,1)&&\textrm{d}.\quad \color{red}(2,1)\\ \textrm{b}.\quad (-1,2)&\textrm{c}.\quad (1,2)&\textrm{e}.\quad (-2,-1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos (\theta _{1}+\theta _{2}) & -\sin (\theta _{1}+\theta _{2})\\ \sin (\theta _{1}+\theta _{2}) & \cos (\theta _{1}+\theta _{2}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos (75^{\circ}+15^{\circ})& -\sin (75^{\circ}+15^{\circ})\\ \sin (75^{\circ}+15^{\circ}) & \cos (75^{\circ}+15^{\circ}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ}&-\sin 90^{\circ}\\ \sin 90^{\circ}&\cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 1 \end{pmatrix} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 12.&\textrm{Jika garis}\: \: 3x-2y+5=0\: \: \textrm{dicerminkan }\\ &\textrm{terhadap garis}\: \: y=-x\: \: \textrm{kemudian}\\ &\textrm{didilatasikan dengan pusat (1,-2) }\\ &\textrm{dengan faktor skala 2, maka persamaan}\\ & \textrm{bayangannya adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x-2y-10=0&&\\ \textrm{b}.\quad x+2y-10=0&\\ \textrm{c}.\quad x-6y+5=0&\\ \textrm{d}.\quad x+2y-12=0\\ \textrm{e}.\quad \color{red}2x-3y+18=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Proses}&\: \textrm{untuk refleksinya}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&-1\\ -1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ -x \end{pmatrix}\\ \textrm{proses}&\: \textrm{dilatasinya}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}\begin{pmatrix} x'-1\\ y'+2 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-2\\ 2y'+4 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-1\\ 2y'+2 \end{pmatrix}\\ &=\begin{pmatrix} 2(-y)-1\\ 2(-x)+2 \end{pmatrix}\\ &\begin{cases} x &=-\displaystyle \frac{1}{2}(y''-2) \\ y &=-\displaystyle \frac{1}{2}(x''+1) \end{cases} \end{aligned}\\ &\begin{aligned}\textrm{Sehingga persam}&\textrm{aan bayangan}\\ \textrm{garisnya adalah}:&\\ 3x&-2y+5=0\\ 3\left ( -\displaystyle \frac{1}{2}(y''-2) \right )&-2\left ( -\displaystyle \frac{1}{2}(x''+1) \right )+5=0\\ -\displaystyle \frac{3}{2}y''+3 &+(x''+1)+5=0\\ 2x&-3y+6+2+10=0\\ 2x&-3y+18=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 13.&\textrm{Titik A(4,-4) dicerminkan terhadap}\\ &\textrm{garis}\: \: y=x\tan 15^{\circ}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: A'(a,b)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}&&\textrm{d}.\quad \color{red}4\sqrt{3}\\ \textrm{b}.\quad 2\sqrt{3}&\textrm{c}.\quad 3\sqrt{3}&\textrm{e}.\quad 6\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} a\\ b \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.15^{\circ}& \sin 2.15^{\circ}\\ \sin 2.15^{\circ} & -\cos 2.15^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \cos 30^{\circ}&\sin 30^{\circ}\\ \sin 30^{\circ}&-\cos 30^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2}\sqrt{3} & \displaystyle \frac{1}{2}\\ \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} 2\sqrt{3}-2\\ 2+2\sqrt{3} \end{pmatrix}\\ &\begin{cases} a &=2\sqrt{3}-2 \\ b &=2+2\sqrt{3} \end{cases}\\ \textrm{mak}&\textrm{a nilai dari}\\ a+b&=\left ( 2\sqrt{3}-2+2+2\sqrt{3} \right )\\ &=4\sqrt{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 14.&\textrm{Lingkaran}\: \: x^{2}+y^{2}-5x+8y+7=0\\ & \textrm{ditranslasikan oleh}\: \: T=\begin{pmatrix} m\\ n \end{pmatrix}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: x^{2}+y^{2}-9x+2y+6=0.\\ & \textrm{Nilai}\: \: m+n=\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad \color{red}5\\ \textrm{b}.\quad 3&\qquad\textrm{c}.\quad 4\qquad&\textrm{e}.\quad 6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui sebuah lingkaran dengan persamaan}:\\ & \color{blue}x^{2}+y^{2}-5x+8y+7=0\\ \textrm{kar}&\textrm{ena akibat translasi, maka}\\ &\begin{cases} x & =x'-m \\ y & =y'-n \end{cases}\\ &x^{2}+y^{2}-5x+8y+7=0\\ \textrm{seh}&\textrm{ingga}\\ &\Leftrightarrow \color{purple}(x'-m)^{2}+(y'-n)^{2}-5(x'-m)+8(y'-n)+7=0\\ &\Leftrightarrow \color{purple}x'^{2}+y'^{2}-2mx'-2ny'+m^{2}+n^{2}-5x'+5m+8y'-8n+7=0\\ &\Leftrightarrow \color{purple}x'^{2}+y'^{2}-(2m+5)x'+(8-2n)y'+m^{2}+n^{2}+5m-8n+7=0\\ &\qquad \equiv \: \color{purple}x'^{2}+y'^{2}-9x'+2y'+6=0\qquad (\color{black}\textbf{akhir bayangan})\\ &\begin{cases} 9 &=2m+5 \Rightarrow m=2\\ 2 & =8-2n \: \Rightarrow \, \: n=3 \end{cases}\\ \textrm{Jad}&\textrm{i , nilai}\: \: m+n=2+3=5\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 15.&\textrm{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-\displaystyle \frac{1}{3}x\sqrt{3}\: ,\: \textrm{maka bayangan dari}\\ &\textrm{titik \textit{A} tersebut adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&&\\ \textrm{b}.\quad \color{red}A'\left ( -1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left (-1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( -1+\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ y&=-\displaystyle \frac{1}{3}x\sqrt{3}=\left ( -\displaystyle \frac{1}{3}\sqrt{3} \right )x\\ &=\left (-\tan 30^{\circ} \right )x=\tan \left ( 180^{\circ}-30^{\circ} \right )x\\ &=\tan 150^{\circ}.x\\ \textrm{maka}\: \: \theta &=150^{\circ}\quad \Rightarrow \quad 2\theta =300^{\circ}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\color{purple}\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} \cos 300^{\circ} & \sin 300^{\circ} \\ \sin 300^{\circ} & -\cos 300^{\circ} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3}\\ -\displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\begin{pmatrix} -1-\displaystyle \frac{1}{2}\sqrt{3}\\ \sqrt{3}-\displaystyle \frac{1}{2} \end{pmatrix} \end{aligned} \end{array}$
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