Exercise Zone : Refleksi (Pencerminan)

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

Tipe:

Standar SBMPTN HOTS

No.

Hasil pencerminan titik (-2,2) terhadap garis {x+y=0} dilanjutkan pencerminan terhadap garis {x-y=0} adalah....

Ganesha Operation

(-2,2)\longrightarrow(-2,2)\longrightarrow(2,-2)

Jadi, hasil pencerminannya adalah (2,-2).

No.

Parabola y=ax^2+bx+c, puncaknya (p,q) dicerminkan terhadap garis {y=1} menghasilkan {y=x^2-x-6}. Milai {a+b+c+p+q} adalah

1. \dfrac{67}4
2. \dfrac{33}4
3. 8

4. \dfrac{31}4
5. \dfrac{30}4

Jika A direfleksikan terhadap T menghasilkan A', maka jika A' direfleksikan terhadap T akan menghasilkan A. Sehingga bisa dikatakan bahwa {y=ax^2+bx+c} adalah bayangan {y=x^2-x-6} setelah dicerminkan terhadap garis {y=1}.
$$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}x\\ 2(1)-y\end{array}\right)\\ & =\left(\begin{array}{c}x\\ 2-y\end{array}\right)\end{array}$$

x'=x
y'=2-y\longrightarrow y=2-y'

$$\begin{array}{rl}y& =x^2-x-6\\ 2-y^{\prime }& ={\left(x^{\prime }\right)}^2-x^{\prime }-6\\ -y^{\prime }& ={\left(x^{\prime }\right)}^2-x^{\prime }-6-2\\ -y^{\prime }& ={\left(x^{\prime }\right)}^2-x^{\prime }-8\\ y^{\prime }& =-{\left(x^{\prime }\right)}^2+x^{\prime }+8\\ y& =-x^2+x+8\end{array}$$

a=-1, b=1, c=8

$$\begin{array}{rl}p& =-{\displaystyle \frac{b}{2a}}\\ & =-{\displaystyle \frac{1}{2(-1)}}\\ & ={\displaystyle \frac{1}{2}}\end{array}$$

$$\begin{array}{rl}q& =-p^2+p+8\\ & =-{\left({\displaystyle \frac{1}{2}}\right)}^2+{\displaystyle \frac{1}{2}}+8\\ & =-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}+8\\ & ={\displaystyle \frac{33}{4}}\end{array}$$

$$\begin{array}{rl}a+b+c+p+q& =-1+1+8+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{33}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{67}{4}}}}}}\end{array}$$

Jadi, a+b+c+p+q=\dfrac{67}4. JAWAB: A

No.

Hasil pencerminan lingkaran {x^2 + y^2 + x-2y-4 = 0} terhadap garis {x =-2} dan {y = 1} adalah

1. {x^2 + y^2-2x-2y + 15 = 0}
2. {x^2 + y^2 + 7x-2y + 8 = 0}
3. {x^2 + y^2-7x + 2y-15 = 0}

4. {x^2 + y^2 + 2x-7y-16 = 0}
5. {x^2 + y^2-2x + 7y-16 = 0}

SKMP Agustus 2020

$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}2(-2)-x\\ 2(1)-y\end{array}\right)\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}-4-x\\ 2-y\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}-4-x^{\prime }\\ 2-y^{\prime }\end{array}\right)\end{array}$

$\begin{array}{rl}{x}^2+y^2+x-2y-4& =0\\ {(-4-x^{\prime })}^2+{(2-y^{\prime })}^2+(-4-x^{\prime })-2(2-y^{\prime })-4& =0\\ 16+8x^{\prime }+{\left(x^{\prime }\right)}^2+4-4y^{\prime }+{\left(y^{\prime }\right)}^2-4-x^{\prime }-4+2y^{\prime }-4& =0\\ {\left(x^{\prime }\right)}^2+{\left(y^{\prime }\right)}^2+7x^{\prime }-2y^{\prime }+8& =0\\ x^2+y^2+7x-2y+8& =0\end{array}$

Jadi, hasil pencerminannya adalah { x^2+y^2+7x-2y+8=0}. JAWAB: B

No.

Titik A(-1,3) refleksikan terhadap garis x=2 menghasilkan titik ....

1. A'(-5,3)
2. A'(-3,3)
3. A'(1,3)

4. A'(3,3)
5. A'(5,3)

Grup Telegram

$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}2(2)-(-1)\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}5\\ 3\end{array}\right)\end{array}$

Jadi, A'(5,3). JAWAB: E

No.

Parabola {y=ax^2+bx+c}, puncaknya {(p,q)} dicerminkan terhadap garis {y=-1} menghasilkan {y=x^2+x-2}. Nilai {a+b+c+p+q} adalah

1. -\dfrac92
2. -\dfrac94
3. -2

4. \dfrac94
5. \dfrac92

Bisa dikatakan bahwa {y=ax^2+bx+c} adalah bayangan {y=x^2+x-2} setelah dicerminkan terhadap garis {y=-1}.
x'=x\rightarrow x=x'
y'=2(-1)-y=-2-y\rightarrow y=-2-y'

$$\begin{array}{rl}y& =x^2+x-2\\ -2-y^{\prime }& =(x^{\prime }{)}^2+x^{\prime }-2\\ -y^{\prime }& =(x^{\prime }{)}^2+x^{\prime }\\ y^{\prime }& =-(x^{\prime }{)}^2-x^{\prime }\\ y& =-x^2-x\end{array}$$
a=-1, b=-1, c=0

$$\begin{array}{rl}p& =-{\displaystyle \frac{b}{2a}}\\ & =-{\displaystyle \frac{-1}{2(-1)}}\\ & ={\displaystyle \frac{1}{2}}\end{array}$$

$$\begin{array}{rl}q& =-p^2-p\\ & =-{\left({\displaystyle \frac{1}{2}}\right)}^2-{\displaystyle \frac{1}{2}}\\ & =-{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2}}\\ & =-{\displaystyle \frac{3}{4}}\end{array}$$

$$\begin{array}{rl}a+b+c+p+q& =-1+(-1)+0+{\displaystyle \frac{1}{2}}+(-{\displaystyle \frac{3}{4}})\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{9}{4}}}}}}\end{array}$$

Jadi, {a+b+c+p+q=-\dfrac94}. JAWAB: B

No.

Garis 12y - 7x = 84 direfleksikan terhadap titik pusat O(0,0), maka bayangannya adalah

Grup Telegram

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

$\begin{array}{rl}12y-7x& =84\\ 12(-y^{\prime })-7(-x^{\prime })& =84\\ -12y^{\prime }+7x^{\prime }& =84\\ 7x^{\prime }-12y^{\prime }& =84\\ 7x-12y& =84\end{array}$

Jadi, bayangannya adalah 7x-12y=84

No.

Titik B(-17, 9) direfleksikan terhadap garis {y=-8}, maka bayangannya adalah

Grup Telegram

B(-17, 9)\rightarrow B'(-17,2(-8)-9)=B'(-17,-25)

Jadi, bayangannya adalah B'(-17,-25)

No.

Titik A(-2,-5) dicerminkan terhadap titik O kemudian dilanjutkan dengan pencerminan terhadap sumbu x. Tentukan bayangan titik A tersebut.

BSE Kelas XI

Refleksi terhadap titik O

$$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}-2\\ -5\end{array}\right)\\ & =\left(\begin{array}{c}2\\ 5\end{array}\right)\end{array}$$

Refleksi terhadap sumbu x

$$\begin{array}{rl}\left(\begin{array}{c}x"\\ y"\end{array}\right)& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}2\\ 5\end{array}\right)\\ & =\left(\begin{array}{c}2\\ -5\end{array}\right)\end{array}$$

Jadi, bayangan titik A adalah A"(2,-5).

No.

Garis 2x - y + 5 = 0 dicerminkan terhadap titik O(0,0) kemudian dilanjutkan dengan pencerminan terhadap sumbu y. Tentukan persamaan bayangan garis tersebut.

BSE Kelas XI

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

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

$$\begin{array}{rl}2x-y+5& =0\\ 2x^{\prime }-(-y^{\prime })+5& =0\\ 2x^{\prime }+y^{\prime }+5& =0\\ 2x+y+5& =0\end{array}$$

Jadi, persamaan bayangan garis tersebut adalah 2x+y+5=0.

No.

Titik A(-1, -3) dicerminkan terhadap titik O(0, 0) kemudian dilanjutkan dengan pencerminan terhadap sumbu y dan dilanjutkan lagi dengan pencerminan terhadap garis y = x. Tentukan bayangan titik A tersebut.

BSE Kelas XI

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

Jadi, bayangan titik A tersebut adalah A'(3,-1)