Exercise Zone : Transformasi Geometri

Berikut ini adalah kumpulan soal mengenai Transformasi Geometri. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

---

No.

Jika suatu garis {l:x+2y=4} dirotasikan 45\degree searah jarum jam dengan pusat rotasi titik asal (0,0) kemudian dilanjutkan dengan dicerminkan terhadap sumbu y, maka diperoleh persamaan garis {g:ax+by=c}. Nilai {a+b+c=} ....

1. 3-\dfrac23\sqrt3
2. 1+\dfrac23\sqrt3
3. \dfrac23\sqrt3

4. 1-\dfrac23\sqrt3
5. 3+\dfrac23\sqrt3

SUMBER

$\begin{array}{rl}{T}_1& =\left(\begin{array}{cc}\cos (-{45}^{\circ })& -\sin (-{45}^{\circ })\\ \sin (-{45}^{\circ })& \cos (-{45}^{\circ })\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {45}^{\circ }& \sin {45}^{\circ }\\ -\sin {45}^{\circ }& \cos {45}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\\ -{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)\end{array}$

$T_2=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$

$\begin{array}{rl}\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}{cc}{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\\ -{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}-{\displaystyle \frac{1}{2}}\sqrt{2}& -{\displaystyle \frac{1}{2}}\sqrt{2}\\ -{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\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}-{\displaystyle \frac{1}{2}}\sqrt{2}& -{\displaystyle \frac{1}{2}}\sqrt{2}\\ -{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)}^{-1}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & ={\displaystyle \frac{1}{(-{\displaystyle \frac{1}{2}}\sqrt{2})\left({\displaystyle \frac{1}{2}}\sqrt{2}\right)-(-{\displaystyle \frac{1}{2}}\sqrt{2})(-{\displaystyle \frac{1}{2}}\sqrt{2})}}\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\\ {\displaystyle \frac{1}{2}}\sqrt{2}& -{\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & ={\displaystyle \frac{1}{-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}}}\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\\ {\displaystyle \frac{1}{2}}\sqrt{2}& -{\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & ={\displaystyle \frac{1}{-1}}\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\\ {\displaystyle \frac{1}{2}}\sqrt{2}& -{\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & =-1\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\\ {\displaystyle \frac{1}{2}}\sqrt{2}& -{\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & =\left(\begin{array}{cc}-{\displaystyle \frac{1}{2}}\sqrt{2}& -{\displaystyle \frac{1}{2}}\sqrt{2}\\ -{\displaystyle \frac{1}{2}}\sqrt{2}& {\displaystyle \frac{1}{2}}\sqrt{2}\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & =\left(\begin{array}{c}-{\displaystyle \frac{1}{2}}\sqrt{2}{x}^{\prime }-{\displaystyle \frac{1}{2}}\sqrt{2}{y}^{\prime }\\ -{\displaystyle \frac{1}{2}}\sqrt{2}{x}^{\prime }+{\displaystyle \frac{1}{2}}\sqrt{2}{y}^{\prime }\end{array}\right)\end{array}$
x=-\dfrac12\sqrt2x'-\dfrac12\sqrt2y'
y=-\dfrac12\sqrt2x'+\dfrac12\sqrt2y'

$\begin{array}{rl}x+2y& =4\\ -{\displaystyle \frac{1}{2}}\sqrt{2}{x}^{\prime }-{\displaystyle \frac{1}{2}}\sqrt{2}{y}^{\prime }+2(-{\displaystyle \frac{1}{2}}\sqrt{2}{x}^{\prime }+{\displaystyle \frac{1}{2}}\sqrt{2}{y}^{\prime })& =4\\ -{\displaystyle \frac{1}{2}}\sqrt{2}{x}^{\prime }-{\displaystyle \frac{1}{2}}\sqrt{2}{y}^{\prime }-\sqrt{2}{x}^{\prime }+\sqrt{2}{y}^{\prime }& =4\\ -{\displaystyle \frac{3}{2}}\sqrt{2}{x}^{\prime }+{\displaystyle \frac{1}{2}}\sqrt{2}{y}^{\prime }& =4\end{array}$

a= -\dfrac32\sqrt2
b=\dfrac12\sqrt2
c=4

$\begin{array}{rl}a+b+c& =-{\displaystyle \frac{3}{2}}\sqrt{2}+{\displaystyle \frac{1}{2}}\sqrt{2}+4\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4-{\displaystyle \frac{1}{2}}\sqrt{2}}}}}\end{array}$

Jadi, {a+b+c=4-\dfrac12\sqrt2}.

No.

Bayangan garis {x+2y+3=0} oleh transformasi yang bersesuaian dengan matriks $$\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$ dilanjutkan oleh rotasi pusat O sejauh 90\degree adalah ....

1. {y+2x+3=0}
2. {-y-2x+3=0}
3. {y+2x-3=0}

4. {-y+2x+3=0}
5. {y-2x+3=0}

$$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}y\\ x\end{array}\right)\end{array}$$

x'=y\rightarrow y=x'
y'=x\rightarrow x=y'

$$\begin{array}{rl}x+2y+3& =0\\ y^{\prime }+2x^{\prime }+3& =0\\ y+2x+3& =0\end{array}$$

Jadi, bayangan garis {x+2y+3=0} oleh transformasi yang bersesuaian dengan matriks $$\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$ dilanjutkan oleh rotasi pusat O sejauh 90\degree adalah {y+2x+3=0}. JAWAB: A

No.

Jika titik (s,t) dirotasi sejauh 270\degree berlawanan arah jarum jam terhadap titik pusat, kemudian dicerminkan terhadap y=t diperoleh titik (-2,3-t), maka s+3t=

1. 1
2. 2
   
3. 3
   

4. 4
5. 5
   

$$\begin{array}{rl}\left(\begin{array}{c}{s}^{\prime }\\ t^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 270{}^{\circ }& -\sin 270{}^{\circ }\\ \sin 270{}^{\circ }& \cos 270{}^{\circ }\end{array}\right)\left(\begin{array}{c}s\\ t\end{array}\right)\\ & =\left(\begin{array}{cc}0& -(-1)\\ -1& 0\end{array}\right)\left(\begin{array}{c}s\\ t\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{c}s\\ t\end{array}\right)\\ & =\left(\begin{array}{c}t\\ -s\end{array}\right)\end{array}$$

$$\begin{array}{rl}\left(\begin{array}{c}s"\\ t"\end{array}\right)& =\left(\begin{array}{c}t\\ 2t-(-s)\end{array}\right)\\ \left(\begin{array}{c}-2\\ 3-t\end{array}\right)& =\left(\begin{array}{c}t\\ 2t+s\end{array}\right)\end{array}$$

$$\begin{array}{rl}3-t& =2t+s\\ 3& =3t+s\\ s+3t& =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$$

Jadi, s+3t=3. JAWAB: C

No.

Persamaan garis hasil pencerminan dari {3𝑥 - 5𝑦 + 15 = 0} dengan garis {y = x} dilanjutkan dengan tranlasi $$\left(\begin{array}{c}2\\ 3\end{array}\right)$$ adalah ...

1. 3y+5x-14 = 0
2. 3y -5x+14 = 0
3. 5x -3y +14 = 0

4. 5x +3y -14 = 0
5. 5x -3x - 14 = 0

SUMBER

$$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}2\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}y\\ x\end{array}\right)+\left(\begin{array}{c}2\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}y+2\\ x+3\end{array}\right)\end{array}$$
x'=y+2\rightarrow y=x'-2
y'=x+2\rightarrow x=y'-3

Bayangannya,
\begin{aligned} 3𝑥 - 5𝑦 + 15 &= 0\ 3(y'-3)-5(x'-2)+15&=0\ 3y'-9-5x'+10+15&=0\ -5x'+3y'+16&=0\ 5x'-3y'-16&=0\ 5x-3y-16&=0

Jadi, persamaan garis hasil pencerminan dari {3𝑥 - 5𝑦 + 15 = 0} dengan garis {y = x} dilanjutkan dengan tranlasi $$\left(\begin{array}{c}2\\ 3\end{array}\right)$$ adalah 5x-3y-16=0. JAWAB: -

No.

Tentukan koordinat titik A, jika titik dirotasi sejauh 90\degree berlawanan arah jarum jam terhadap titik pusat O(0,0), kemudian dicerminkan terhadap y = b diperoleh titik (5, 2-b)

1. (5, 13)
2. (5. -13)
3. (13, 5)

4. (-13, 5)
5. (-13,-5)

SUMBER

Rotasi 90\degree

$$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{c}-b\\ a\end{array}\right)\end{array}$$

Refleksi y=b

$$\begin{array}{rl}\left(\begin{array}{c}x"\\ y"\end{array}\right)& =\left(\begin{array}{c}-b\\ 2b-a\end{array}\right)\\ \left(\begin{array}{c}5\\ 2-b\end{array}\right)& =\left(\begin{array}{c}-b\\ 2b-a\end{array}\right)\end{array}$$

$$\begin{array}{rl}5& =-b\\ b& =-5\end{array}$$

$$\begin{array}{rl}2-b& =2b-a\\ 2-(-5)& =2(-5)-a\\ 2+5& =-10-a\\ 7& =-10-a\\ a& =-10-7\\ & =-17\end{array}$$

Jadi, A(-17,-5). JAWAB: -