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

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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

\(\eqalign{ T_1&=\pmatrix{\cos(-45^\circ)&-\sin(-45^\circ)\\\sin(-45^\circ)&\cos(-45^\circ)}\\ &=\pmatrix{\cos45^\circ&\sin45^\circ\\-\sin45^\circ&\cos45^\circ}\\ &=\pmatrix{\dfrac12\sqrt2&\dfrac12\sqrt2\\-\dfrac12\sqrt2&\dfrac12\sqrt2}\\ }\)

\(T_2=\pmatrix{-1&0\\0&1}\)

\(\eqalign{ \pmatrix{x'\\y'}&=\pmatrix{-1&0\\0&1}\pmatrix{\dfrac12\sqrt2&\dfrac12\sqrt2\\-\dfrac12\sqrt2&\dfrac12\sqrt2} \pmatrix{x\\y}\\ \pmatrix{x'\\y'}&=\pmatrix{-\dfrac12\sqrt2&-\dfrac12\sqrt2\\-\dfrac12\sqrt2&\dfrac12\sqrt2}\pmatrix{x\\y}\\ \pmatrix{x\\y}&=\pmatrix{-\dfrac12\sqrt2&-\dfrac12\sqrt2\\-\dfrac12\sqrt2&\dfrac12\sqrt2}^{-1}\pmatrix{x'\\y'}\\ &=\dfrac1{\left(-\dfrac12\sqrt2\right)\left(\dfrac12\sqrt2\right)-\left(-\dfrac12\sqrt2\right)\left(-\dfrac12\sqrt2\right)}\pmatrix{\dfrac12\sqrt2&\dfrac12\sqrt2\\\dfrac12\sqrt2&-\dfrac12\sqrt2}\pmatrix{x'\\y'}\\ &=\dfrac1{-\dfrac12-\dfrac12}\pmatrix{\dfrac12\sqrt2&\dfrac12\sqrt2\\\dfrac12\sqrt2&-\dfrac12\sqrt2}\pmatrix{x'\\y'}\\ &=\dfrac1{-1}\pmatrix{\dfrac12\sqrt2&\dfrac12\sqrt2\\\dfrac12\sqrt2&-\dfrac12\sqrt2}\pmatrix{x'\\y'}\\ &=-1\pmatrix{\dfrac12\sqrt2&\dfrac12\sqrt2\\\dfrac12\sqrt2&-\dfrac12\sqrt2}\pmatrix{x'\\y'}\\ &=\pmatrix{-\dfrac12\sqrt2&-\dfrac12\sqrt2\\-\dfrac12\sqrt2&\dfrac12\sqrt2}\pmatrix{x'\\y'}\\ &=\pmatrix{-\dfrac12\sqrt2x'-\dfrac12\sqrt2y'\\-\dfrac12\sqrt2x'+\dfrac12\sqrt2y'}\\ }\)
x=-\dfrac12\sqrt2x'-\dfrac12\sqrt2y'
y=-\dfrac12\sqrt2x'+\dfrac12\sqrt2y'

\(\eqalign{ x+2y&=4\\ -\dfrac12\sqrt2x'-\dfrac12\sqrt2y'+2\left(-\dfrac12\sqrt2x'+\dfrac12\sqrt2y'\right)&=4\\ -\dfrac12\sqrt2x'-\dfrac12\sqrt2y'-\sqrt2x'+\sqrt2y'&=4\\ -\dfrac32\sqrt2x'+\dfrac12\sqrt2y'&=4 }\)

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

\(\eqalign{ a+b+c&=-\dfrac32\sqrt2+\dfrac12\sqrt2+4\\ &=\boxed{\boxed{4-\dfrac12\sqrt2}} }\)

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

No.

Bayangan garis {x+2y+3=0} oleh transformasi yang bersesuaian dengan matriks \begin{pmatrix}1&0\\0&-1\end{pmatrix} 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{aligned} \begin{pmatrix}x'\\y'\end{pmatrix}&=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\\ &=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\\ &=\begin{pmatrix}y\\x\end{pmatrix} \end{aligned}

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

\begin{aligned} x+2y+3&=0\\ y'+2x'+3&=0\\ y+2x+3&=0 \end{aligned}

Jadi, bayangan garis {x+2y+3=0} oleh transformasi yang bersesuaian dengan matriks \begin{pmatrix}1&0\\0&-1\end{pmatrix} 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{aligned} \begin{pmatrix}s'\\t'\end{pmatrix}&=\begin{pmatrix}\cos270\degree&-\sin270\degree\\\sin270\degree&\cos270\degree\end{pmatrix}\begin{pmatrix}s\\t\end{pmatrix}\\ &=\begin{pmatrix}0&-(-1)\\-1&0\end{pmatrix}\begin{pmatrix}s\\t\end{pmatrix}\\ &=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}s\\t\end{pmatrix}\\ &=\begin{pmatrix}t\\-s\end{pmatrix} \end{aligned}

\begin{aligned} \begin{pmatrix}s"\\t"\end{pmatrix}&=\begin{pmatrix}t\\2t-(-s)\end{pmatrix}\\ \begin{pmatrix}-2\\3-t\end{pmatrix}&=\begin{pmatrix}t\\2t+s\end{pmatrix} \end{aligned}

\begin{aligned} 3-t&=2t+s\\ 3&=3t+s\\ s+3t&=\boxed{\boxed{3}} \end{aligned}

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

No.

Persamaan garis hasil pencerminan dari {3𝑥 - 5𝑦 + 15 = 0} dengan garis {y = x} dilanjutkan dengan tranlasi \begin{pmatrix}2\\3\end{pmatrix} 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{aligned} \begin{pmatrix}x'\\y'\end{pmatrix}&=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}2\\3\end{pmatrix}\\ &=\begin{pmatrix}y\\x\end{pmatrix}+\begin{pmatrix}2\\3\end{pmatrix}\\ &=\begin{pmatrix}y+2\\x+3\end{pmatrix} \end{aligned}
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 \begin{pmatrix}2\\3\end{pmatrix} 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{aligned} \begin{pmatrix}x'\\y'\end{pmatrix}&=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}\\ &=\begin{pmatrix}-b\\a\end{pmatrix} \end{aligned}

Refleksi y=b

\begin{aligned} \begin{pmatrix}x"\\y"\end{pmatrix}&=\begin{pmatrix}-b\\2b-a\end{pmatrix}\\ \begin{pmatrix}5\\2-b\end{pmatrix}&=\begin{pmatrix}-b\\2b-a\end{pmatrix}\end{aligned}

\begin{aligned} 5&=-b\\ b&=-5\end{aligned}

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

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