$\begin{array}{ll}\\ 11.&\textrm{Garis}\: \: 2x+y+4=0\: \: \textrm{ditranslasikan}\\ &\textrm{oleh}\: \: \begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{dilanjutkan transformasi} \\ &\textrm{oleh}\: \: \begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\: \: \textrm{persamaan bayangannya}\\ &\textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2x+y+3=0&&\\ \textrm{b}.\quad 2x-3y+3=0&\\ \textrm{c}.\quad 2x+3y+3=0&\\ \textrm{d}.\quad 3x+2y+3=0\\ \textrm{e}.\quad 3x-2y+3=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}\: &=\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} -2\\ 5 \end{pmatrix}=\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x-2+2y+10\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x+2y+8\\ y+5 \end{pmatrix}\\ \textrm{Diper}&\textrm{oleh}\\ &\begin{array}{lllllllll}\\ \quad x''&=x+2y+8\\ \: \: \: 2y''&=2y+10\qquad\quad-\\\hline x''-2y''&=x-2\\ \Leftrightarrow \quad x&=x''-2y''+2\: \color{black}....(1)\\ \textrm{maka}\\ \qquad y&=y''-5\quad \qquad....(2) \end{array}\\ \textrm{sehin}&\textrm{gga}\\ 2x+&y+4=0\\ 2(x''&-2y''+2)+(y''-5)+4=0\\ 2x''-&3y''+3=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ 2x-&3y+3=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 12.&\textrm{Titik A(4,-4) dicerminkan terhadap}\\ &\textrm{garis}\: \: y=x\tan 15^{\circ}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: A'(a,b)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}&&\textrm{d}.\quad 4\sqrt{3}\\ \textrm{b}.\quad 2\sqrt{3}&\textrm{c}.\quad 3\sqrt{3}&\textrm{e}.\quad 6\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} a\\ b \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.15^{\circ}& \sin 2.15^{\circ}\\ \sin 2.15^{\circ} & -\cos 2.15^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \cos 30^{\circ}&\sin 30^{\circ}\\ \sin 30^{\circ}&-\cos 30^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2}\sqrt{3} & \displaystyle \frac{1}{2}\\ \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} 2\sqrt{3}-2\\ 2+2\sqrt{3} \end{pmatrix}\\ &\begin{cases} a &=2\sqrt{3}-2 \\ b &=2+2\sqrt{3} \end{cases}\\ \textrm{mak}&\textrm{a nilai dari}\\ a+b&=\left ( 2\sqrt{3}-2+2+2\sqrt{3} \right )\\ &=4\sqrt{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 13.&\textrm{Lingkaran}\: \: x^{2}+y^{2}-5x+8y+7=0\\ & \textrm{ditranslasikan oleh}\: \: T=\begin{pmatrix} m\\ n \end{pmatrix}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: x^{2}+y^{2}-9x+2y+6=0.\\ & \textrm{Nilai}\: \: m+n=\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 3&\qquad\textrm{c}.\quad 4\qquad&\textrm{e}.\quad 6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui sebuah lingkaran dengan persamaan}:\\ & x^{2}+y^{2}-5x+8y+7=0\\ \textrm{kar}&\textrm{ena akibat translasi, maka}\\ &\begin{cases} x & =x'-m \\ y & =y'-n \end{cases}\\ &x^{2}+y^{2}-5x+8y+7=0\\ \textrm{seh}&\textrm{ingga}\\ &\Leftrightarrow (x'-m)^{2}+(y'-n)^{2}-5(x'-m)+8(y'-n)+7=0\\ &\Leftrightarrow x'^{2}+y'^{2}-2mx'-2ny'+m^{2}+n^{2}-5x'+5m+8y'-8n+7=0\\ &\Leftrightarrow x'^{2}+y'^{2}-(2m+5)x'+(8-2n)y'+m^{2}+n^{2}+5m-8n+7=0\\ &\qquad \equiv \: x'^{2}+y'^{2}-9x'+2y'+6=0\qquad (\textbf{akhir bayangan})\\ &\begin{cases} 9 &=2m+5 \Rightarrow m=2\\ 2 & =8-2n \: \Rightarrow \, \: n=3 \end{cases}\\ \textrm{Jad}&\textrm{i , nilai}\: \: m+n=2+3=5\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 14.&\textrm{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-\displaystyle \frac{1}{3}x\sqrt{3}\: ,\: \textrm{maka bayangan dari}\\ &\textrm{titik A tersebut adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&&\\ \textrm{b}.\quad A'\left ( -1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left (-1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( -1+\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ y&=-\displaystyle \frac{1}{3}x\sqrt{3}=\left ( -\displaystyle \frac{1}{3}\sqrt{3} \right )x\\ &=\left (-\tan 30^{\circ} \right )x=\tan \left ( 180^{\circ}-30^{\circ} \right )x\\ &=\tan 150^{\circ}.x\\ \textrm{maka}\: \: \theta &=150^{\circ}\quad \Rightarrow \quad 2\theta =300^{\circ}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 300^{\circ} & \sin 300^{\circ} \\ \sin 300^{\circ} & -\cos 300^{\circ} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3}\\ -\displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\begin{pmatrix} -1-\displaystyle \frac{1}{2}\sqrt{3}\\ \sqrt{3}-\displaystyle \frac{1}{2} \end{pmatrix} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 15.&\textrm{Bayangan titik A(2,4) dicerminkan }\\ &\textrm{terhadap garis}\: \: y-x=0\: \: \textrm{dilanjutkan}\\ &\textrm{ke garis}\: \: x\sqrt{3}-3y=0\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad A'\left ( 2+\sqrt{3},-1+2\sqrt{3} \right )&&\\ \textrm{b}.\quad A'\left ( 2+\sqrt{3},1-2\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left ( 1-\sqrt{3},-2+\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( -2+\sqrt{3},1+2\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( 2-\sqrt{3},1-2\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ &\begin{cases} x\sqrt{3}-3y=0 & \Leftrightarrow y=\displaystyle \frac{1}{3}\sqrt{3}x\\ &\Leftrightarrow y=\tan 30^{\circ}.x\\\\ x-y=0 & \Leftrightarrow y=x \end{cases}\\\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.30^{\circ} & \sin 2.30^{\circ} \\ \sin 2.30^{\circ} & -\cos 2.30^{\circ} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}\sqrt{3}\\ \displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} \sqrt{3}+2\\ -1+2\sqrt{3} \end{pmatrix} \end{aligned} \end{array}$.
Tidak ada komentar:
Posting Komentar
Informasi