Linear transformation of a line Question:
Find the image of the line $y=5x+9$ by rotation with center $O(0,0)$ with rotation angle of $270^{\circ}$ followed  by dilatation at the center $O(0,0)$ with a scale factor of $3$.
My Attempt:
Given the line $y=5x+9$, it passes through $(-1.8,0)$ and $(0,9)$. I used these two points to perform the rotation of $270^{\circ}$ about the origin. When rotated anti-clockwise, the image of these two points becomes $(0,1.8)$ and $(9,0)$ respectively. With a dilatation at the center with a scale factor of $3$, these points becomes $(0, 5.4)$ and $(27,0)$, hence the image of the line $y=5x+9$ becomes $y=-0.2x+5.4$ or $10y=-2x+54$.
When rotated clockwise, the image of the line $y=5x+9$ becomes $y=-0.2x-5.4$ or $10y=-2x-54$.
I would like to check if the answer/ approach is correct for this question? If not correct, how should I approach it.
Thank you.
 A: This is correct. You can verify the answer by using linear algebra as well. The original line is given by
$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0\\9 \end{pmatrix}+ t \begin{pmatrix} 1 \\ 5 \end{pmatrix}; (t\in\mathbb{R})$$
The rotation by $270$° represented by the matrix R, and the dilation by 3 represented by the matrix D are given by:
$$R=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}; D=\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$$
Finding the transformed line is equivalent to performing the following operation:
$$\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\bigg(\begin{pmatrix} 0\\9 \end{pmatrix}+ t \begin{pmatrix} 1 \\ 5 \end{pmatrix}\bigg)$$
This results in the following equation:
$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 27\\0 \end{pmatrix}+ t \begin{pmatrix} 15 \\ -3 \end{pmatrix}$$
Which can be written as this system
$$x=27+15t$$
$$y=-3t$$
solving for $y(x)$ yields the same answer you obtained.
