Contoh Soal 3 Transformasi Geometri

$\begin{array}{l}\\ 11.& \text{Titik A(1,-2) dirotasikan sejauh}{15}^{\circ }\\ & \text{kemudian dilanjutkan}{75}^{\circ }\text{dengan pusat }\\ & O(0,0)\text{maka bayangan akhir titik A adalah}...\\ & \begin{array}{l}\\ \text{a}.(-2,1)& & \text{d}.(2,1)\\ \text{b}.(-1,2)& \text{c}.(1,2)& \text{e}.(-2,-1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos ({\theta }_1+{\theta }_2)& -\sin ({\theta }_1+{\theta }_2)\\ \sin ({\theta }_1+{\theta }_2)& \cos ({\theta }_1+{\theta }_2)\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos ({75}^{\circ }+{15}^{\circ })& -\sin ({75}^{\circ }+{15}^{\circ })\\ \sin ({75}^{\circ }+{15}^{\circ })& \cos ({75}^{\circ }+{15}^{\circ })\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ 1\end{array}\right)\end{array}\end{array}$.

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

$\begin{array}{l}\\ 13.& \text{Titik A(4,-4) dicerminkan terhadap}\\ & \text{garis}y=x\tan {15}^{\circ }\text{menghasilkan}\\ & \text{bayangan}{A}^{\prime }(a,b)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\sqrt{3}& & \text{d}.4\sqrt{3}\\ \text{b}.2\sqrt{3}& \text{c}.3\sqrt{3}& \text{e}.6\sqrt{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}a\\ b\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.15}^{\circ }& \sin {2.15}^{\circ }\\ \sin {2.15}^{\circ }& -\cos {2.15}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {30}^{\circ }& \sin {30}^{\circ }\\ \sin {30}^{\circ }& -\cos {30}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{c}2\sqrt{3}-2\\ 2+2\sqrt{3}\end{array}\right)\\ & \{\begin{array}{ll}a& =2\sqrt{3}-2\\ b& =2+2\sqrt{3}\end{array}\\ \text{mak}& \text{a nilai dari}\\ a+b& =(2\sqrt{3}-2+2+2\sqrt{3})\\ & =4\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Lingkaran}{x}^2+y^2-5x+8y+7=0\\ & \text{ditranslasikan oleh}T=\left(\begin{array}{c}m\\ n\end{array}\right)\text{menghasilkan}\\ & \text{bayangan}{x}^2+y^2-9x+2y+6=0.\\ & \text{Nilai}m+n=...\\ & \begin{array}{l}\\ \text{a}.2& & \text{d}.5\\ \text{b}.3& \text{c}.4& \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui sebuah lingkaran dengan persamaan}:\\ & x^2+y^2-5x+8y+7=0\\ \text{kar}& \text{ena akibat translasi, maka}\\ & \{\begin{array}{ll}x& =x^{\prime }-m\\ y& =y^{\prime }-n\end{array}\\ & x^2+y^2-5x+8y+7=0\\ \text{seh}& \text{ingga}\\ & \iff (x^{\prime }-m{)}^2+(y^{\prime }-n{)}^2-5(x^{\prime }-m)+8(y^{\prime }-n)+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-2m{x}^{\prime }-2n{y}^{\prime }+m^2+n^2-5x^{\prime }+5m+8y^{\prime }-8n+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-(2m+5)x^{\prime }+(8-2n)y^{\prime }+m^2+n^2+5m-8n+7=0\\ & \equiv x^{\prime 2}+y^{\prime 2}-9x^{\prime }+2y^{\prime }+6=0(\mathbf{\text{akhir bayangan}})\\ & \{\begin{array}{ll}9& =2m+5\Rightarrow m=2\\ 2& =8-2n\Rightarrow n=3\end{array}\\ \text{Jad}& \text{i , nilai}m+n=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-{\displaystyle \frac{1}{3}x\sqrt{3},\text{maka bayangan dari}}\\ & \text{titik textit{A} tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& & \\ \text{b}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& \\ \text{c}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})& \\ \text{d}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})\\ \text{e}.A^{\prime }(-1+{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ y& =-{\displaystyle \frac{1}{3}x\sqrt{3}=(-{\displaystyle \frac{1}{3}\sqrt{3}})x}\\ & =(-\tan {30}^{\circ })x=\tan ({180}^{\circ }-{30}^{\circ })x\\ & =\tan {150}^{\circ }.x\\ \text{maka}\theta & ={150}^{\circ }\Rightarrow 2\theta ={300}^{\circ }\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {300}^{\circ }& \sin {300}^{\circ }\\ \sin {300}^{\circ }& -\cos {300}^{\circ }\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\\ -{\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}-1-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \sqrt{3}-{\displaystyle \frac{1}{2}}\end{array}\right)\end{array}\end{array}$