Contoh Soal 5 Transformasi Geometri

$\begin{array}{l}\\ 21.& \text{Jika setiap titik pada grafik dengan}\\ & \text{dengan persamaan}y=\sqrt{x}\text{dicerminkan}\\ & \text{terhadap garis}y=x,\text{maka persamaan}\\ & \text{grafik yang dihasilkan adalah}...\\ & \begin{array}{l}\\ \text{a}.y=x^2,x\ge 0& & \\ \text{b}.y=-\sqrt{x},x\ge 0& \\ \text{c}.y=-x^2,x\le 0& \\ \text{d}.y=\sqrt{-x},x\le 0\\ \text{e}.y=-\sqrt{-x},x\le 0\end{array}\\ \\ & \mathbf{\text{UMB Tahun 2011 Kode 152}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ y& =\sqrt{x},\text{atau}{y}^2=x\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 1}}\\ \text{mak}& \text{a}\text{saat dicerminkan terhadap}\\ \text{gari}& \text{s}y=x,\text{adalah}{x}^2=y\\ \text{atau}& y=x^2.\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 2}}\\ \text{Jika}& \text{ingin dikerjakan dengan rumus}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =M_{x=y}\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}y\\ x\end{array}\right)\\ \text{Sela}& \text{njutnya hasilnya disubstitusikan}\\ \text{ke p}& \text{ersamaan}y=\sqrt{x}\Rightarrow x^{\prime }=\sqrt{y^{\prime }}\\ \sqrt{y^{\prime }}& =x^{\prime }\text{maka}\\ y^{\prime }& ={\left(x^{\prime }\right)}^2\text{selanjutnya}\\ y& =x^2\end{array}\end{array}$.

Sebelum dicerminkan terhadap garis y=x

Gambar kurva/grafik setelah cerminkan terhadap garis y=x

$\begin{array}{l}\\ 22.& \text{Transformasi}T\text{adalah pencerminan}\\ & \text{terhadap garis}y={\displaystyle \frac{x}{3}\text{dilanjutkan oleh}}\\ & \text{pencerminan terhadap garis}y=-3x.\\ & \text{Matriks yang bersesuian dengan}\\ & \text{transformasi}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)& \\ \text{c}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& \\ \text{d}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ \text{e}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{SBMPTN Tahun 2013 Kode 433}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \text{sebu}& \text{ah persamaan garis lurus}\\ \text{dapa}& \text{t dituliskan dengan}:y=mx\\ \text{Dike}& \text{tahui pula bahwa ada 2 garis}:\\ y_1& ={\displaystyle \frac{1}{3}x\text{dan}{y}_2=-3x}\\ \text{seba}& \text{gai representasi transformasi}T.\\ \text{Kare}& \text{na}{m}_1\times m_2=\left({\displaystyle \frac{1}{3}}\right)(-3)=-1\\ \text{bera}& \text{rti 2 garis di atas saling tegak}\\ \text{luru}& \text{s dan hal ini seperti rotasi 2}\\ \text{kali}& {90}^{\circ }\text{atau}{180}^{\circ }\\ \text{Jadi},& T=\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ \iff & T=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
3. Sharma,S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.

Contoh Soal 4 Transformasi Geometri

$\begin{array}{l}\\ 16.& \text{Bayangan titik A(2,4) dicerminkan }\\ & \text{terhadap garis}y-x=0\text{dilanjutkan}\\ & \text{ke garis}x\sqrt{3}-3y=0\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(2+\sqrt{3},-1+2\sqrt{3})& & \\ \text{b}.A^{\prime }(2+\sqrt{3},1-2\sqrt{3})& \\ \text{c}.A^{\prime }(1-\sqrt{3},-2+\sqrt{3})& \\ \text{d}.A^{\prime }(-2+\sqrt{3},1+2\sqrt{3})\\ \text{e}.A^{\prime }(2-\sqrt{3},1-2\sqrt{3})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ & \{\begin{array}{ll}x\sqrt{3}-3y=0& \iff y={\displaystyle \frac{1}{3}\sqrt{3}x}\\ & \iff y=\tan {30}^{\circ }.x\\ \\ x-y=0& \iff y=x\end{array}\\ \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.30}^{\circ }& \sin {2.30}^{\circ }\\ \sin {2.30}^{\circ }& -\cos {2.30}^{\circ }\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}\\ {\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ 4\end{array}\right)\\ & =\left(\begin{array}{c}\sqrt{3}+2\\ -1+2\sqrt{3}\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Jika}{T}_1=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\text{dan}{T}_2=\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\\ & \text{maka bayangan garis}x+y+1=0\\ & \text{oleh}{T}_2\circ T_1\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.x-2y-1=0& & \\ \text{b}.x+2y-1=0& \\ \text{c}.x+2y+1=0& \\ \text{d}.x-2y+1=0\\ \text{e}.x+y-1=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T_2\circ T_1\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2+5& -4+5\\ -1+3& -2+3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}3& 1\\ 2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}3x+y\\ 2x+y\end{array}\right)\\ \text{Dipe}& \text{roleh}\\ & \begin{array}{l}\\ x^{\prime }& =3x+y\\ y^{\prime }& =2x+y-\\ x^{\prime }-y^{\prime }& =x\\ \iff x& =x^{\prime }-y^{\prime }....(1)\\ \text{maka}\\ y& =x^{\prime }-3x\\ & =x^{\prime }-3(x^{\prime }-y^{\prime })\\ & =3y^{\prime }-2x^{\prime }....(2)\end{array}\\ \text{Sehin}& \text{gga}\\ x+y& +1=0\\ x^{\prime }-& y^{\prime }+3y^{\prime }-2x^{\prime }+1=0\\ -x^{\prime }+& 2y^{\prime }+1=0\\ x^{\prime }-& 2y^{\prime }-1=0\\ \text{maka}& \text{bayangan garisnya}\\ x-2& y-1=0\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Garis}2x+y+4=0\text{ditranslasikan}\\ & \text{oleh}\left(\begin{array}{c}-2\\ 5\end{array}\right)\text{dilanjutkan transformasi}\\ & \text{oleh}\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\text{persamaan bayangannya}\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.2x+y+3=0& & \\ \text{b}.2x-3y+3=0& \\ \text{c}.2x+3y+3=0& \\ \text{d}.3x+2y+3=0\\ \text{e}.3x-2y+3=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}-2\\ 5\end{array}\right)=\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ \left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)& =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x-2+2y+10\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x+2y+8\\ y+5\end{array}\right)\\ \text{Diper}& \text{oleh}\\ & \begin{array}{l}\\ x^{″}& =x+2y+8\\ 2y^{″}& =2y+10-\\ x^{″}-2y^{″}& =x-2\\ \iff x& =x^{″}-2y^{″}+2....(1)\\ \text{maka}\\ y& =y^{″}-5....(2)\end{array}\\ \text{sehin}& \text{gga}\\ 2x+& y+4=0\\ 2(x^{″}& -2y^{″}+2)+(y^{″}-5)+4=0\\ 2x^{″}-& 3y^{″}+3=0\\ \text{maka}& \text{bayangan garisnya}\\ 2x-& 3y+3=0\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Diketahui}M\text{adalah pencerminan terhadap}\\ & \text{garis}y=-x\text{dan}T\text{adalah transformasi}\\ & \text{yang dinyatakan oleh matriks}\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\\ & \text{Koordinat bayangan titik}A(2,-8)\text{oleh}\\ & \text{transformasi}M\text{dilanjutkan oleh}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-10,2)& & \\ \text{b}.(-2,-10)& \\ \text{c}.(10,2)& \\ \text{d}.(-10,-2)\\ \text{e}.(2,10)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T\circ M\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}0-3& -2+0\\ 0+1& 0+0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}-3& -2\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{c}-6+16\\ 2+0\end{array}\right)\\ & =\left(\begin{array}{c}10\\ 2\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Jika}W\text{adalah transformasi oleh}\\ & \text{matriks}\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{maka titik mula}\\ & \text{dari}{W}^{\prime }(-2,5)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-11,-2)& & \\ \text{b}.(11,-2)& \\ \text{c}.(-2,11)& \\ \text{d}.(2,11)\\ \text{e}.(12,11)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dimi}& \text{salkan}:\\ A& =\left(\begin{array}{c}-2\\ 5\end{array}\right),\text{dan}\\ W& =\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{serta}X=\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{mak}& \text{a}\\ & \begin{array}{|c|}\hline \begin{array}{rl}A& =BX\\ B^{-1}A& =B^{-1}BX\\ B^{-1}A& =I.X\\ B^{-1}A& =X\\ X& =B^{-1}A\end{array}\\ \hline\end{array}\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\displaystyle \frac{1}{\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|}\left(\begin{array}{cc}1& 0\\ -3& 1\end{array}\right)\left(\begin{array}{c}-2\\ 5\end{array}\right)}\\ & =1.\left(\begin{array}{c}-2+0\\ 6+5\end{array}\right)\\ & =\left(\begin{array}{c}-2\\ 11\end{array}\right)\end{array}\end{array}$

Contoh Soal 3 Transformasi Geometri

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

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

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

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

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

Contoh Soal 2 Transformasi Geometri

$\begin{array}{l}\\ 6.& \text{Bayangan untuk titik P(2,5) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{A}}(1,3)\text{sejauh}{180}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(1,0)& & \text{d}.(2,0)\\ \text{b}.(0,1)& \text{c}.(0,2)& \text{e}.(1,2)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Karena rotasi de}& \text{ngan pusat A sebesar}{180}^{\circ },\\ \text{maka}& \\ R(A(1,3),{180}^{\circ })& =\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{sehingga bayang}& \text{an titik P(2,5)-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}2-1\\ 5-3\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ -2\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Bayangan kurva}xy=6\text{oleh rotasi sebesar}\\ & {\displaystyle \frac{\pi }{2}\text{dengan pusat}O(0,0)\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.xy=-6& & \text{d}.x(y-x)=6\\ \text{b}.xy=6& & \text{e}.x(x+y)=-6\\ \text{c}.x(x-y)=6& \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Karena rotasi d}& \text{engan pusat O sebesar}\\ {\displaystyle \frac{\pi }{2}={90}^{\circ },\text{maka}}& \\ R(O(0,0),{90}^{\circ })& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ \text{sehingga bayan}& \text{gan semua titik yang }\\ \text{terletak pada k}& \text{urva adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-y\\ x\end{array}\right)\\ & \{\begin{array}{ll}x& =y^{\prime }\\ y& =-x^{\prime }\end{array}\\ \text{Selanjunya}\text{unt}& \text{uk bayangan kurvanya }\\ \text{adalah}:& \\ xy& =6\\ y^{\prime }.(-x^{\prime })& =6\\ x^{\prime }{y}^{\prime }& =-6\\ \text{Jadi , persamaa}& \text{n kurva bayangannya}\\ \text{adalah}& xy=-6\end{array}\end{array}$.

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

$\begin{array}{l}\\ 9.& \text{Jika}{M}_x\text{adalah pencerminan terhadap sumbu-X }\\ & \text{dan}{M}_{y=x}\text{adalah pencerminan terhadap garis}\\ & y=x,\text{maka matriks transformasi tunggal }\\ & \text{yang mewakili}{M}_x\circ M_{y=x}=....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)& & \text{d}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{b}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)& & \text{e}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ \text{c}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ M_{y=x}& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\end{array}\\ M_x\circ M_{y=x}& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ & =\left(\begin{array}{cc}0+0& 1+0\\ 0-1& 0+0\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Diketahui vektor}\overrightarrow{x}\text{dirotasikan terhadap titik asal}\\ & O\text{sebesar}\theta >0\text{searah jarum jam}.\\ & \text{Kemudian hasilnya dicerminkan terhadap garis}y=0\\ & \text{menghasilkan vektor}\overrightarrow{y}.\\ & \text{Jika}\overrightarrow{y}=A.\overrightarrow{x},\text{maka matriks}A-\text{nya adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)& & \\ \text{c}.\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{d}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\\ \text{e}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui bah}\text{wa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ R_{-\theta }& =\left(\begin{array}{cc}\cos (-\theta )& -\sin (-\theta )\\ \sin (-\theta )& \cos (-\theta )\end{array}\right)=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\\ & A=M_x\circ R_{-\theta }=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\end{array}$.

Contoh Soal 1 Transformasi Geometri

$\begin{array}{l}\\ 1.& \text{Suatu translasi yang memetakan titik P(9,8) }\\ & \text{ke titik}{\text{P}}^{\prime }(14,-2)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{c}5\\ -10\end{array}\right)& & \text{d}.\left(\begin{array}{c}6\\ 6\end{array}\right)\\ \text{b}.\left(\begin{array}{c}5\\ 6\end{array}\right)& \text{c}.\left(\begin{array}{c}23\\ -10\end{array}\right)& \text{e}.\left(\begin{array}{c}5\\ 2\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\text{T}+\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{T}& =\left(\begin{array}{c}{x}^{\prime }-x\\ y^{\prime }-y\end{array}\right)\\ & =\left(\begin{array}{c}14-9\\ -2-8\end{array}\right)\\ & =\left(\begin{array}{c}5\\ -10\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ & \text{Maka bayangan titik M}(2,-1)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(7,10)& & \text{d}.(1,10)\\ \text{b}.(10,7)& \text{c}.(1,4)& \text{e}.(4,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ \begin{array}{c}x=2\\ y=-1\end{array}\}& \Rightarrow \{\begin{array}{ll}{x}^{\prime }& =2(2)+3(-1)=4-3=1\\ y^{\prime }& =3(2)+2(-1)=6-2=4\end{array}\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 5.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \text{Yang merupakan titik invarian (tidak berubah) }\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(0,0)& & \text{d}.(0,-1)\\ \text{b}.(1,-1)& \text{c}.(1,0)& \text{e}.(1,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \begin{array}{|cccc|}\hline \text{NO}& \text{Titik}& \begin{array}{rl}& \text{Disubstitusikan ke}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\end{array}& \begin{array}{rl}& \text{Keterangan}\\ & \text{Titik}\end{array}\\ \text{a}.& (0,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(0)-4=-4\end{array}& \text{Varian}\\ \text{b}& (1,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(-1)-4=-1\end{array}& \mathbf{\text{Invarian}}\\ \text{c}& (1,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(0)-4=-2\end{array}& \text{Varian}\\ \text{d}& (0,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(-1)-4=-3\end{array}& \text{Varian}\\ \text{e}& (1,1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(1)-4=-3\end{array}& \text{Varian}\\ \hline\end{array}\end{array}\end{array}$

Transformasi Geometri (XI Matematika Wajib)

 A. Pengertian 

Transformasi Geometri adalah suatu perubahan objek geometri atau suatu pemetaan dari suatu titik-titik ke himpunan titik-titik yang lain pada bidang kartesius.

Dari pengertian di atas jelas bahwa aturan transformasi sebagaimana fungsi atau pemetaan dan transformasi ini selanjutnya dapat disimbolkan dengan sebuah huruf kapital, misal M, T, R, dan lain sebagainya. Sebagai misal titik P(x,y) oleh transformasi T menghasilkan titik baru yaitu P'(x',y') dan operasi ini dapat dituliskan dengan:

$\begin{array}{rl}& P(x,y)\stackrel{T}{\to }{P}^{\prime }(x^{\prime },y^{\prime })\end{array}$.

B. Matriks Transformasi

Misalkan suatu transfomasi T memetakan sebuah titik A(x,y) ke A'(x',y') 

selanjutnya perhatikan ilustrasi berikut:

$\boxed{{\displaystyle \begin{array}{rl}A(x,y)& \underset{.}{\overset{Transformasi=T}{\to }}{A}^{\prime }(x^{\prime },y^{\prime })=A^{\prime }(ax+by,cx+dy)\\ \\ \Rightarrow & \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\underset{\underset{transformasi}{Matriks}}{\underbrace{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)}}\left(\begin{array}{c}x\\ y\end{array}\right)\end{array}}}$.

C. Jenis-Jenis Transformasi dengan matriks yang sesuaian

1. Translasi (Geseran)

$\begin{array}{|lcc|}\hline \begin{array}{rl}& \text{Jenis}\\ & \text{Transformasi}\end{array}& \text{Rumus}& \text{Matriks}\\ \text{Translasi}& (x,y)\underset{.}{\overset{\left(\begin{array}{c}a\\ b\end{array}\right)}{\to }}(x+a,y+b)& \left(\begin{array}{c}a\\ b\end{array}\right)\\ \hline\end{array}$.

2. Rotasi (Perputaran)

$\begin{array}{rl}& \begin{array}{|lcc|}\hline \begin{array}{rl}& \text{Jenis}\\ & \text{Transformasi}\end{array}& \text{Rumus}& \text{Matriks}\\ \text{Rotasi}& & \\ \begin{array}{rl}& \text{Pusat rotasi}\\ & [O,\alpha ]\end{array}& \begin{array}{rl}& \{\begin{array}{l}{x}^{\prime }=...\\ y^{\prime }=...\end{array}\\ & \begin{array}{rl}& \text{Lihat}\\ & \text{di bawah}\\ & \text{tulisan}\\ & \text{warna}\\ & \text{biru}\end{array}\end{array}& \left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)\\ \begin{array}{rl}& \text{Pusat}\\ & (a,b)\text{sudut}\alpha \end{array}& \left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)=& \begin{array}{rl}& \text{Lihat}\\ & \text{di bawah}\\ & \text{tulisan}\\ & \text{warna}\\ & \text{merah}\end{array}\\ \hline\end{array}\\ & \{\begin{array}{l}{x}^{\prime }=x\cos \alpha -y\sin \alpha \\ y^{\prime }=x\sin \alpha +y\cos \alpha \end{array}\\ & ▹▹▹▹\left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right).\left(\begin{array}{c}x-a\\ y-b\end{array}\right)\end{array}$.

3. Refleksi (Pencerminan)

$\begin{array}{|lcc|}\hline \text{Refleksi}& & \\ \text{terhadap sumbu}-\text{X}& (x,y)\to (x,-y)& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ \text{terhadap sumbu}-\text{Y}& (x,y)\to (-x,y)& \left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ \text{terhadap garis y = x}& (x,y)\to (y,x)& \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ \text{terhadap garis y = -x}& (x,y)\to (-y,-x)& \left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\\ \text{terhadap garis x = h}& (x,y)\to (2h-x,y)& \left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}2h\\ 0\end{array}\right)\\ \text{terhadap garis y = x}& (x,y)\to (y,x)& \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ \text{terhadap garis y = -x}& (x,y)\to (-y,-x)& \left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\\ \text{terhadap garis x = h}& (x,y)\to (2h-x,y)& \left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}2h\\ 0\end{array}\right)\\ \text{terhadap garis y = k}& (x,y)\to (x,2k-y)& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}0\\ 2k\end{array}\right)\\ \text{pusat}(0,0)\{\begin{array}{l}y=mx\\ m=\tan \alpha \end{array}& & \left(\begin{array}{cc}\cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{array}\right)\\ \hline\end{array}$.

4. Dilatasi (Perkalian)

$\begin{array}{rl}& \begin{array}{|lcc|}\hline \begin{array}{rl}& \text{Jenis}\\ & \text{Transformasi}\end{array}& \text{Rumus}& \text{Matriks}\\ \text{Dilatasi}& & \\ \text{Pusat}[O,k]& (x,y)\to (kx,ky)& \left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)\\ \begin{array}{rl}& \text{Pusat}(a,b)\\ & \text{faktor skala}k\end{array}& \left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)& \begin{array}{rl}& \text{Lihat}\\ & \text{di bawah}\\ & \text{tulisan}\\ & \text{warna}\\ & \text{merah}\end{array}\\ \begin{array}{rl}& \text{Luas bangun}\\ & \text{ datar}\end{array}& \text{Misal bangun A}& \text{T}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ & {\mathbf{\text{Bangun A}}}^{\prime }& =\text{det T}\times \text{A}\\ \hline\end{array}\\ & ▹▹▹▹▹▹\left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)=\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right).\left(\begin{array}{c}x-a\\ y-b\end{array}\right)\end{array}$.

Catatan:

Translasi, refleksi, dan rotasi suatu objek adalah bagian dari transformasi yang hanya mengubah posisi objek saja, sehingga jenis transformasi-transformasi ini juga disebut dengan transformasi isometri

D. Bayangan Kurva dan Komposisi Transformasi

$\begin{array}{|ll|}\hline \text{Bayangan Kurva}y=f(x)& \text{Komposisi Transformasi}\\ \begin{array}{rl}\text{Lan}& \text{gkah-langkah}:\\ 1.& \text{Tentukan bayangan titiknya}\\ & (x,y)\to (x^{\prime },y^{\prime })\\ 2.& \text{Salanjutnya tentukan}x\text{dan}y\\ & \text{dalam}{x}^{\prime }\text{dan}{y}^{\prime }\\ 3.& \text{Substitusikan}x\text{dan}y\\ & \text{ke}y=f(x)\end{array}& \begin{array}{rl}\text{Lan}& \text{gkah-langkah}:\\ 1.& \text{Selesaikan sesuai urutan transformasi}\\ & (x,y)\underset{.}{\overset{T_1}{\to }}(x^{\prime },y^{\prime })\underset{.}{\overset{T_2}{\to }}(x^{″},y^{″})\\ 2.& \text{Jika dapat disederhanakan kedua transformasi}\\ & \text{tersebut di atas, maka cukup dengan}\\ & (x,y)\underset{.}{\overset{T_2\circ T_1}{\to }}(x^{″},y^{″})\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah bayangan dari segitiga PQR dengan}\\ & P(0,4),Q(-1,1),\text{dan}R(3,6).\\ & \text{oleh translasi}T=\left(\begin{array}{c}5\\ -2\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}\\ & \{\begin{array}{ll}\left(\begin{array}{c}{x}_P^{{}^{\prime }}\\ y_P^{{}^{\prime }}\end{array}\right)& =T+\left(\begin{array}{c}{x}_P\\ y_P\end{array}\right)=\left(\begin{array}{c}5\\ -2\end{array}\right)+\left(\begin{array}{c}0\\ 4\end{array}\right)=\left(\begin{array}{c}5+0\\ -2+4\end{array}\right)=\left(\begin{array}{c}5\\ 2\end{array}\right)\\ \left(\begin{array}{c}{x}_Q^{{}^{\prime }}\\ y_Q^{{}^{\prime }}\end{array}\right)& =\cdots \text{isilah sendiri}\\ \left(\begin{array}{c}{x}_R^{{}^{\prime }}\\ y_R^{{}^{\prime }}\end{array}\right)& =\cdots \text{isilah sendiri}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah bayangan dari garis}y=2x+4\\ & \text{oleh translasi}T=\left(\begin{array}{c}-1\\ 2\end{array}\right).\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|cc|}\hline \mathbf{\text{Bayangan Titik-titik}}& \mathbf{\text{Bayangan Garis}}\\ \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T+\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ 2\end{array}\right)+\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}-1+x\\ 2+y\end{array}\right)\\ & \{\begin{array}{ll}{x}^{\prime }& =-1+x\iff x=x^{\prime }+1\\ y^{\prime }& =2+y\iff y=y^{\prime }-2\end{array}\end{array}& \begin{array}{rl}y& =2x+4\\ y^{\prime }-2& =2(x^{\prime }+1)+4\\ y^{\prime }& =2x+2+4+2\\ & =2x+8\\ \text{Jadi},& \mathbf{\text{bayangan garisnya}}\\ \text{adala}& \text{h}:\\ y& =2x+8\\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah bayangan titik A(4,6) oleh rotasi yang berpusat }\\ & \text{di titik P(3,-2) dengan sudut putar sebesar}{90}^{\circ }\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Untuk Ro}& \text{tasi yang berpusat di}(a,b)\text{dengan sudut}\alpha \text{adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\left(\begin{array}{c}4-3\\ 6-(-2)\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 8\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}-8\\ 1\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}-5\\ -1\end{array}\right)\\ \text{Jadi},& \text{bayangan titik A adalah}{\text{A}}^{\prime }(-5,-1)\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah bayangan titik A(4,6) }\\ & \text{oleh dilatasi yang berpusat di titik P(3,-2)}\\ & \text{dengan faktor skala}k=2\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Bayangan}& \text{titik A-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}4-3\\ 6-(-2)\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}1\\ 8\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ 16\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}5\\ 14\end{array}\right)\\ \text{Jadi},& \text{bayangan titik A-nya adalah}{\text{A}}^{\prime }(5,14)\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah bayangan titik A(4,6) }\\ & \text{oleh translasi}t\text{dilanjutkan}s\text{dengan}\\ & \text{matriks transformasi berturut-turut }\\ & \text{adalah}T=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\text{dan}S=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Bayangan}& \text{titik A-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =S\times T\times \left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\left(\begin{array}{c}4\\ 6\end{array}\right)\\ & =\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)\left(\begin{array}{c}4\\ 6\end{array}\right)\\ & =\left(\begin{array}{c}26\\ 16\end{array}\right)\\ \text{Jadi},& \text{bayangan titik A-nya adalah}{\text{A}}^{\prime }(26,16)\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Suatu kurva}y={}^3\log (2x-2)\text{memiliki bayangan}\\ & y={}^3\log \left({\displaystyle \frac{2x+3}{3}}\right)\text{oleh translasi}\\ & T=\left(\begin{array}{c}a\\ b\end{array}\right).\text{Tentukanlah nilai}a+b\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}\\ y& ={}^3\log (2x-2)\iff 3^y=2x-2(\text{benda})\\ y& ={}^3\log \left({\displaystyle \frac{2x+3}{3}}\right)\\ & \iff 3^y=\left({\displaystyle \frac{2x+3}{3}}\right)(\mathbf{\text{bayangan}})\\ \text{sehingga}& \text{untuk bayangan}\\ 3^{y^{\prime }-b}& =2(x^{\prime }-a)-2\iff 3^{y^{\prime }}{.3}^{-b}=2(x^{\prime }-a)-2\\ & \iff 3^{y^{\prime }}={\displaystyle \frac{2(x^{\prime }-a)-2}{3^{-b}}={\displaystyle \frac{2x^{\prime }+3}{3}}}\\ \text{Jadi},& \{\begin{array}{ll}a& ={\displaystyle \frac{5}{2}}\\ b& =-1\end{array}\\ \text{Sehingga}& a+b={\displaystyle \frac{5}{2}+(-1)={\displaystyle \frac{3}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tentukanlah bayangan garis}ax+by+c=0\text{oleh transformasi}\\ & \text{yang bersesuaian dengan matriks}\left(\begin{array}{cc}1& -2\\ 3& -4\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|cc|}\hline \mathbf{\text{Proses Awal}}& \mathbf{\text{Penentuan Bayangan}}\\ \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}1& -2\\ 3& -4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\left(\begin{array}{cc}1& -2\\ 3& -4\end{array}\right)}^{-1}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & ={\displaystyle \frac{1}{\left|\begin{array}{cc}1& -2\\ 3& -4\end{array}\right|}\left(\begin{array}{cc}-4& 2\\ -3& 1\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)}\\ & ={\displaystyle \frac{1}{-4+6}\left(\begin{array}{c}-4x^{\prime }+2y^{\prime }\\ -3x^{\prime }+y^{\prime }\end{array}\right)}\\ & ={\displaystyle \frac{1}{2}\left(\begin{array}{c}-4x^{\prime }+2y^{\prime }\\ -3x^{\prime }+y^{\prime }\end{array}\right)}\\ & \{\begin{array}{ll}x& =-2x^{\prime }+y^{\prime }\\ y& =-{\displaystyle \frac{3}{2}{x}^{\prime }+{\displaystyle \frac{1}{2}{y}^{\prime }}}\end{array}\end{array}& \begin{array}{rl}ax+by+c& =0\\ a(-2x^{\prime }+y^{\prime })+b(-{\displaystyle \frac{3}{2}{x}^{\prime }+\frac{1}{2}{y}^{\prime }})+c& =0\\ -2a{x}^{\prime }-{\displaystyle \frac{3}{2}b{x}^{\prime }+a{y}^{\prime }+{\displaystyle \frac{1}{2}b{y}^{\prime }+c}}& =0\\ (-4a-3b)x^{\prime }+(2a+b)y^{\prime }+2c& =0\\ & \\ \mathbf{\text{Jadi, bayangan garisnya adalah}}:& \\ & \\ (-4a-3b)x+(2a+b)y+2c& =0\\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Diketahui kurva}y=4x^2-9\text{dicerminkan terhadap sumbu-X kemudian}\\ & \text{ditranslasikan dengan}\left(\begin{array}{c}-1\\ 2\end{array}\right).\text{Ordinat titik potong terhadap sumbu-Y adalah}....\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}-1\\ 2\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ 2\end{array}\right)+\left(\begin{array}{c}x\\ -y\end{array}\right)\\ & =\left(\begin{array}{c}-1+x\\ 2-y\end{array}\right)\\ & \{\begin{array}{ll}x& =x^{\prime }-1\\ y& =2-y^{\prime }\end{array}\end{array}& \begin{array}{rl}y& =4x^2-9\\ (2-y^{\prime })& =4(x^{\prime }-1{)}^2-9\\ -y^{\prime }& =4(x^{\prime 2}-2x^{\prime }+1)-9-2\\ -y^{\prime }& =4x^{\prime 2}-8x^{\prime }+4-11\\ y^{\prime }& =-4x^{\prime 2}+8x^{\prime }+7\\ & \\ \mathbf{\text{Maka}},& \mathbf{\text{persamaan kurva bayangannya}}:\\ y& =-4x^2+8x+7\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}\text{Sehingga}& \text{ordinat dari titik potong terhadap sumbu-Y-nya adalah}:\\ y& =-4x^2+8x+7,\mathbf{\text{atau}}\\ f(x)& =-4x^2+8x+7\\ f(0)& =-4(0{)}^2+8(0)+7\text{saat}x=0(\text{karena memotong sumbu-Y})\\ & =7\\ \text{Jadi}& \text{ordinatnya adalah}y=f(0)=7\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.