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.