Exercise Zone : Translasi

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

Tipe:

Standar SBMPTN HOTS

No. 1

Diketahui persegi panjang PQRS, P(2,1), Q(5,1), R(5,3), dan S(2,3). Tentukan bayangan bangun tersebut bila ditranslasi dengan

1. T=(4,3)
2. T=(6,-5)

SUMBER

1. T=(4,3)
   $\begin{array}{rl}\left(\begin{array}{cccc}{P}^{\prime }& Q^{\prime }& R^{\prime }& S^{\prime }\end{array}\right)& =\left(\begin{array}{cccc}2& 5& 5& 2\\ 1& 1& 3& 3\end{array}\right)+\left(\begin{array}{cccc}4& 4& 4& 4\\ 3& 3& 3& 3\end{array}\right)\\ & =\left(\begin{array}{cccc}6& 9& 9& 6\\ 4& 4& 6& 6\end{array}\right)\end{array}$
   P'(6,4), Q'(9,4), R'(9,6), dan S'(6,6)
2. T=(6,-5)
   $\begin{array}{rl}\left(\begin{array}{cccc}{P}^{\prime }& Q^{\prime }& R^{\prime }& S^{\prime }\end{array}\right)& =\left(\begin{array}{cccc}2& 5& 5& 2\\ 1& 1& 3& 3\end{array}\right)+\left(\begin{array}{cccc}6& 6& 6& 6\\ -5& -5& -5& -5\end{array}\right)\\ & =\left(\begin{array}{cccc}8& 11& 11& 8\\ -4& -4& -2& -2\end{array}\right)\end{array}$
   P'(8,-4), Q'(11,-4), R'(11,-2), dan S'(8,-2)

Jadi, JAWAB:

No. 2

Titik A(5,-12) ditranslasikan oleh $T=\left(\begin{array}{c}a\\ b\end{array}\right)$ ke titik A'(-4,7). Carilah

1. translasi T
2. hasil translasi persegi panjang ABCD, dengan A(2,1), B(5,4), C(3.6), dan D(0,4) oleh translasi T.

Grup Telegram

1. translasi T
   $\begin{array}{rl}\left(\begin{array}{c}-4\\ 7\end{array}\right)& =\left(\begin{array}{c}5\\ -12\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ \left(\begin{array}{c}a\\ b\end{array}\right)& =\left(\begin{array}{c}-4\\ 7\end{array}\right)-\left(\begin{array}{c}5\\ -12\end{array}\right)\\ T& =\left(\begin{array}{c}-9\\ 19\end{array}\right)\end{array}$
2. hasil translasi persegi panjang ABCD, dengan A(2,1), B(5,4), C(3.6), dan D(0,4) oleh translasi T.
   $\left(\begin{array}{cccc}2& 5& 3& 0\\ 1& 4& 6& 4\end{array}\right)+\left(\begin{array}{cccc}-9& -9& -9& -9\\ 19& 19& 19& 19\end{array}\right)=\left(\begin{array}{cccc}-7& -4& -6& -9\\ 20& 23& 25& 23\end{array}\right)$
   

Jadi, $T=\left(\begin{array}{c}-9\\ 19\end{array}\right)$ A'(-7,20), B'(-4,23), C'(-6.25), dan D'(-9,23).

No. 3

Jika setiap titik pada parabola y=x^2-6 ditranslasikan menurut vektor (3,-2) maka parabola yang dihasilkan adalah ....

1. y=2x^2-6x-4
2. y=2x^2+2x-6
3. y=4x^2-6x+2

4. y=x^2+6x+4
5. y=x^2-6x+1

Ganesha Operation

$$\begin{array}{rl}(y-(-2))& =(x-3{)}^2-6\\ y+2& =x^2-6x+9-6\\ y& =x^2-6x+1\end{array}$$

Jadi, parabola yang dihasilkan adalah y=x^2-6x+1.

No. 4

Diketahui persamaan garis g: 8x-9y+72=0. Tentukan persamaan bayangan garis g oleh:

1. translasi $T=\left(\begin{array}{c}3\\ -4\end{array}\right)$
2. translasi $T=\left(\begin{array}{c}2\\ 6\end{array}\right)$

Grup Telegram

1. $\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}3\\ -4\end{array}\right)$
   $\begin{array}{rl}\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)-\left(\begin{array}{c}3\\ -4\end{array}\right)\\ & =\left(\begin{array}{c}{x}^{\prime }-3\\ y^{\prime }+4\end{array}\right)\end{array}$
   
   $\begin{array}{rl}8x-9y+72& =0\\ 8(x^{\prime }-3)-9(y^{\prime }+4)+72& =0\\ 8x^{\prime }-24-9y^{\prime }-36+72& =0\\ 8x^{\prime }-9y^{\prime }+8& =0\\ 8x-9y+8& =0\end{array}$

2. $\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}2\\ 6\end{array}\right)$
   $\begin{array}{rl}\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)-\left(\begin{array}{c}2\\ 6\end{array}\right)\\ & =\left(\begin{array}{c}{x}^{\prime }-2\\ y^{\prime }-6\end{array}\right)\end{array}$
   
   $\begin{array}{rl}8x-9y+72& =0\\ 8(x^{\prime }-2)-9(y^{\prime }-6)+72& =0\\ 8x^{\prime }-16-9y^{\prime }+54+72& =0\\ 8x^{\prime }-9y^{\prime }+110& =0\\ 8x-9y+110& =0\end{array}$

Jadi, persamaan bayangan garis g adalah: 8x-9y+8=0 8x-9y+110=0

No. 5

Titik P(a,b) dan Q(2,4) ditranslasi oleh $$\left(\begin{array}{c}p\\ q\end{array}\right)$$, hasilnya adalah P'(3,4) dan Q'(a,3b). Nilai dari $$\left(\begin{array}{c}p\\ q\end{array}\right)$$

1. $$\left(\begin{array}{c}2\\ 2\end{array}\right)$$
2. $$\left(\begin{array}{c}2\\ -2\end{array}\right)$$
3. $$\left(\begin{array}{c}-2\\ 2\end{array}\right)$$

4. $$\left(\begin{array}{c}-2\\ 2\end{array}\right)$$
5. $$\left(\begin{array}{c}-2\\ -2\end{array}\right)$$

$$\begin{array}{rl}\left(\begin{array}{c}a\\ b\end{array}\right)+\left(\begin{array}{c}p\\ q\end{array}\right)& =\left(\begin{array}{c}3\\ 4\end{array}\right)\\ \left(\begin{array}{c}p\\ q\end{array}\right)& =\left(\begin{array}{c}3\\ 4\end{array}\right)-\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{c}3-a\\ 4-b\end{array}\right)\end{array}$$

$$\begin{array}{rl}\left(\begin{array}{c}2\\ 4\end{array}\right)+\left(\begin{array}{c}p\\ q\end{array}\right)& =\left(\begin{array}{c}a\\ 3b\end{array}\right)\\ \left(\begin{array}{c}p\\ q\end{array}\right)& =\left(\begin{array}{c}a\\ 3b\end{array}\right)-\left(\begin{array}{c}2\\ 4\end{array}\right)\\ & =\left(\begin{array}{c}a-2\\ 3b-4\end{array}\right)\end{array}$$

$$\left(\begin{array}{c}3-a\\ 4-b\end{array}\right)$$=$$\left(\begin{array}{c}a-2\\ 3b-4\end{array}\right)$$

$$\begin{array}{rl}3-a& =a-2\\ -a-a& =-2-3\\ -2a& =-5\\ a& ={\displaystyle \frac{5}{2}}\end{array}$$

$$\begin{array}{rl}4-b& =3b=4\\ -b-3b& =-4-4\\ -4b& =-8\\ b& =2\end{array}$$

$$\begin{array}{rl}\left(\begin{array}{c}p\\ q\end{array}\right)& =\left(\begin{array}{c}3-a\\ 4-b\end{array}\right)\\ & =\left(\begin{array}{c}3-{\displaystyle \frac{5}{2}}\\ 4-2\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ 2\end{array}\right)}}}}\end{array}$$

Jadi, $$\left(\begin{array}{c}p\\ q\end{array}\right)$$=$$\left(\begin{array}{c}{\displaystyle \frac{1}{2}}\\ 2\end{array}\right)$$. JAWAB: -

No. 6

Kurva y=x^2-6x+9, adalah kurva hasil translasi dari kurva {y=x^2-2x+3} oleh translasi $$\left(\begin{array}{c}a\\ b\end{array}\right)$$. Nilai {a+3b} adalah ....

1. 1
2. 2
3. 5

4. 10
5. 15

$$\begin{array}{rl}y+b& =(x+a{)}^2-6(x+a)+9\\ & =x^2+2ax+a^2-6x-6a+9\\ y& =x^2+(2a-6)x+a^2-6a-b+9\end{array}$$

x^2+(2a-6)x+a^2-6a-b+9=x^2-2x+3

$$\begin{array}{rl}2a-6& =-2\\ 2a& =4\\ a& =2\end{array}$$

$$\begin{array}{rl}{a}^2-6a-b+9& =3\\ 2^2-6(2)-b+9& =3\\ 4-12-b+9& =3\\ -b+1& =3\\ b& =-2\end{array}$$

$$\begin{array}{rl}a+3b& =2+3(-2)\\ & =2-6\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -4}}}}\end{array}$$

Jadi, {a+3b=-4} JAWAB: -