Exercise Zone : Dilatasi

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Tipe:

Standar SBMPTN HOTS

No. 1

Bayangan lingkaran $(x-2{)}^2+(y-1{)}^2=4$ yang di-dilatasi dengan $O(0,0)$ dan faktor skala $-2$ adalah

1. $(2x-2{)}^2+(2y-1{)}^2=4$
2. $(2x-2{)}^2+(2y-1{)}^2=8$

3. ${({\displaystyle \frac{1}{2}}x-2)}^2+{({\displaystyle \frac{1}{2}}y-1)}^2=4$
4. ${({\displaystyle \frac{1}{2}}x-2)}^2+{({\displaystyle \frac{1}{2}}y-1)}^2=-8$

SKMP September 2002

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

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

Jadi, bayangan lingkarannya adalah ${({\displaystyle \frac{1}{2}}x-2)}^2+{({\displaystyle \frac{1}{2}}y-1)}^2=4$.

No. 2

Diketahui $G(0,3)$ jika di dilatasi pada pusat $B(1,2)$ dengan faktor skala $4$. Maka hasil bayangan tersebut adalah.....

Latihan PAS Ganjil 2021

$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =4\left(\begin{array}{c}0-1\\ 3-2\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)\\ & =4\left(\begin{array}{c}-1\\ 1\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}-4\\ 4\end{array}\right)+\left(\begin{array}{c}1\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}-3\\ 6\end{array}\right)\end{array}$

Jadi, bayangannya adalah $G^{\prime }(-3,6)$.

No. 3

Diketahui persamaan awal $6x+9y-30=0$ yang di $D[O,3]$ jika $O$ adalah pusat $(0,0)$ maka tentukan persamaan bayangannya!

Latihan PAS ganjil 2021

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

$x^{\prime }=3x\to x={\displaystyle \frac{x^{\prime }}{3}}$
$y^{\prime }=3y\to y={\displaystyle \frac{y^{\prime }}{3}}$

Bayangannya,
$\begin{array}{rl}6x+9y-30& =0\\ 6\left({\displaystyle \frac{x}{3}}\right)+9\left({\displaystyle \frac{y}{3}}\right)-30& =0\\ 2x+3y-30& =0\end{array}$ }

Jadi, persamaan bayangannya adalah $2x+3y-30=0$.