Latihan Soal 9 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 81.& \text{Bayangan untuk titik A(1,3) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{O}}(0,0)\text{sejauh}{90}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(-1,3)& & \text{d}.(1,-3)\\ \text{b}.(-1,-3)& \text{c}.(-3,1)& \text{e}.(3,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Karena rotasi d}& \text{engan pusat O sebesar}{90}^{\circ },\\ \text{maka}& \\ R(O(0,0),{90}^{\circ })& =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ \text{sehingga}& \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-3\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 82.& \text{Suatu lingkaran dengan jari-jari 4 }\\ & \text{dengan pusat di O(0,0) dtranslasikan}\\ & \text{oleh}\text{T}=\left(\begin{array}{c}2\\ -3\end{array}\right),\text{maka luas }\\ & \text{bayangan lingkaran tersebut adalah}\\ & ....\text{satuan luas}\\ & \begin{array}{l}\\ \text{a}.\pi & & \text{d}.8\pi \\ \text{b}.2\pi & \text{c}.4\pi & \text{e}.16\pi \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui persamaan lingkaran berpusat}\\ & \text{di O dengan}r=4.\text{Karena translasi adalah}\\ & \text{termasuk transformasi isometri(kongruen)}\\ & \text{maka jari-jari lingkaran bayangannya }\\ & \text{akan sama dengan bendanya. Sehingga}\\ & \text{ luas bayangan lingkarannya}\\ & =\pi r^2=\pi \times 4^2=16\pi \end{array}\end{array}$.

$\begin{array}{l}\\ 83.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \text{Yang merupakan titik invarian (tidak berubah) }\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(0,0)& & \text{d}.(0,-1)\\ \text{b}.(1,-1)& \text{c}.(1,0)& \text{e}.(1,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \begin{array}{|cccc|}\hline \text{NO}& \text{Titik}& \begin{array}{rl}& \text{Disubstitusikan ke}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\end{array}& \begin{array}{rl}& \text{Keterangan}\\ & \text{Titik}\end{array}\\ \text{a}.& (0,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(0)-4=-4\end{array}& \text{Varian}\\ \text{b}& (1,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(-1)-4=-1\end{array}& \mathbf{\text{Invarian}}\\ \text{c}& (1,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(0)-4=-2\end{array}& \text{Varian}\\ \text{d}& (0,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(-1)-4=-3\end{array}& \text{Varian}\\ \text{e}& (1,1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(1)-4=-3\end{array}& \text{Varian}\\ \hline\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 84.& \text{Bayangan untuk titik P(2,5) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{A}}(1,3)\text{sejauh}{180}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(1,0)& & \text{d}.(2,0)\\ \text{b}.(0,1)& \text{c}.(0,2)& \text{e}.(1,2)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Karena rotasi de}& \text{ngan pusat A sebesar}{180}^{\circ },\\ \text{maka}& \\ R(A(1,3),{180}^{\circ })& =\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{sehingga bayang}& \text{an titik P(2,5)-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}2-1\\ 5-3\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ -2\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 85.& \text{Bayangan kurva}xy=6\text{oleh rotasi sebesar}\\ & {\displaystyle \frac{\pi }{2}\text{dengan pusat}O(0,0)\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.xy=-6& & \text{d}.x(y-x)=6\\ \text{b}.xy=6& & \text{e}.x(x+y)=-6\\ \text{c}.x(x-y)=6& \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Karena rotasi d}& \text{engan pusat O sebesar}\\ {\displaystyle \frac{\pi }{2}={90}^{\circ },\text{maka}}& \\ R(O(0,0),{90}^{\circ })& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ \text{sehingga bayan}& \text{gan semua titik yang }\\ \text{terletak pada k}& \text{urva adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-y\\ x\end{array}\right)\\ & \{\begin{array}{ll}x& =y^{\prime }\\ y& =-x^{\prime }\end{array}\\ \text{Selanjunya}\text{unt}& \text{uk bayangan kurvanya }\\ \text{adalah}:& \\ xy& =6\\ y^{\prime }.(-x^{\prime })& =6\\ x^{\prime }{y}^{\prime }& =-6\\ \text{Jadi , persamaa}& \text{n kurva bayangannya}\\ \text{adalah}& xy=-6\end{array}\end{array}$.

$\begin{array}{l}\\ 86.& \text{Sebuah lingkaran yang berpusat di (3,4) }\\ & \text{dan menyinggung sumbu-X dicerminkan}\\ & \text{terhadap garis}y=x\text{, maka persamaan }\\ & \text{akhir lingkaran yang terjadi adalah}....\\ & \begin{array}{l}\\ \text{a}.x^2+y^2-8x-6y+9=0& & \\ \text{b}.x^2+y^2+8x+6y+9=0& \\ \text{c}.x^2+y^2+6x+8y+9=0& \\ \text{d}.x^2+y^2-8x-6y+16=0\\ \text{e}.x^2+y^2+8x+6y+16=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Refleksi l}& \text{ingkaran yang berpusat di (3,4) }\\ \text{dan men}& \text{yinggung sumbu-X, }\\ \text{dengan}r& =(y)=4,\\ \text{maka}\text{per}& \text{samaan lingkarannya adalah}:\\ (x-3{)}^2+& (y-4{)}^2=4^2.\text{Karena}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}y\\ x\end{array}\right)\\ & \{\begin{array}{ll}x& =y^{\prime }\\ y& =x^{\prime }\end{array}\\ \text{selanjutn}& \text{ya untuk persamaan bayangan }\\ \text{lingkaran}& \text{nya adalah}:\\ & (y^{\prime }-3{)}^2+(x^{\prime }-4{)}^2=4^2,\\ & \mathbf{\text{menjadi}}\\ & (y-3{)}^2+(x-4{)}^2=4^2,\text{atau}:\\ & x^2+y^2-8x-6y+9=0\end{array}\end{array}$.

$\begin{array}{l}\\ 87.& \text{Jika}{M}_x\text{adalah pencerminan terhadap sumbu-X }\\ & \text{dan}{M}_{y=x}\text{adalah pencerminan terhadap garis}\\ & y=x,\text{maka matriks transformasi tunggal }\\ & \text{yang mewakili}{M}_x\circ M_{y=x}=....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)& & \text{d}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{b}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)& & \text{e}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ \text{c}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ M_{y=x}& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\end{array}\\ M_x\circ M_{y=x}& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ & =\left(\begin{array}{cc}0+0& 1+0\\ 0-1& 0+0\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 88.& \text{Diketahui vektor}\overrightarrow{x}\text{dirotasikan terhadap titik asal}\\ & O\text{sebesar}\theta >0\text{searah jarum jam}.\\ & \text{Kemudian hasilnya dicerminkan terhadap garis}y=0\\ & \text{menghasilkan vektor}\overrightarrow{y}.\\ & \text{Jika}\overrightarrow{y}=A.\overrightarrow{x},\text{maka matriks}A-\text{nya adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)& & \\ \text{c}.\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{d}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\\ \text{e}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui bah}\text{wa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ R_{-\theta }& =\left(\begin{array}{cc}\cos (-\theta )& -\sin (-\theta )\\ \sin (-\theta )& \cos (-\theta )\end{array}\right)=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\\ & A=M_x\circ R_{-\theta }=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 89.& \text{Titik A(1,-2) dirotasikan sejauh}{15}^{\circ }\\ & \text{kemudian dilanjutkan}{75}^{\circ }\text{dengan pusat }\\ & O(0,0)\text{maka bayangan akhir titik A adalah}...\\ & \begin{array}{l}\\ \text{a}.(-2,1)& & \text{d}.(2,1)\\ \text{b}.(-1,2)& \text{c}.(1,2)& \text{e}.(-2,-1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos ({\theta }_1+{\theta }_2)& -\sin ({\theta }_1+{\theta }_2)\\ \sin ({\theta }_1+{\theta }_2)& \cos ({\theta }_1+{\theta }_2)\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos ({75}^{\circ }+{15}^{\circ })& -\sin ({75}^{\circ }+{15}^{\circ })\\ \sin ({75}^{\circ }+{15}^{\circ })& \cos ({75}^{\circ }+{15}^{\circ })\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 90.& \text{Jika garis}3x-2y+5=0\text{dicerminkan }\\ & \text{terhadap garis}y=-x\text{kemudian}\\ & \text{didilatasikan dengan pusat (1,-2) }\\ & \text{dengan faktor skala 2, maka persamaan}\\ & \text{bayangannya adalah}....\\ & \begin{array}{l}\\ \text{a}.x-2y-10=0& & \\ \text{b}.x+2y-10=0& \\ \text{c}.x-6y+5=0& \\ \text{d}.x+2y-12=0\\ \text{e}.2x-3y+18=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Proses}& \text{untuk refleksinya}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-y\\ -x\end{array}\right)\\ \text{proses}& \text{dilatasinya}\\ \left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)& =\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}{x}^{\prime }-1\\ y^{\prime }+2\end{array}\right)+\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2x^{\prime }-2\\ 2y^{\prime }+4\end{array}\right)+\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2x^{\prime }-1\\ 2y^{\prime }+2\end{array}\right)\\ & =\left(\begin{array}{c}2(-y)-1\\ 2(-x)+2\end{array}\right)\\ & \{\begin{array}{ll}x& =-{\displaystyle \frac{1}{2}(y^{″}-2)}\\ y& =-{\displaystyle \frac{1}{2}(x^{″}+1)}\end{array}\end{array}\\ & \begin{array}{rl}\text{Sehingga persam}& \text{aan bayangan}\\ \text{garisnya adalah}:& \\ 3x& -2y+5=0\\ 3(-{\displaystyle \frac{1}{2}(y^{″}-2)})& -2(-{\displaystyle \frac{1}{2}(x^{″}+1)})+5=0\\ -{\displaystyle \frac{3}{2}{y}^{″}+3}& +(x^{″}+1)+5=0\\ 2x& -3y+6+2+10=0\\ 2x& -3y+18=0\end{array}\end{array}$.