Find the Image $2x-3y=0$ given matrix transformation Be $M$ the linear transformation represented by matrix $T$:
$$T = 
    \left[ \begin{matrix}
    5 & 2 \\
    4 & 1 \\
    \end{matrix}\right ]
$$
Mark the correct answer which show the image by $M$ of the line $2x-3y = 0$
(A) $19x - 8y = 0$
(B) $19x - 14y = 0$ (answer)
(C) $19x -15y = 0$
(D) $15x-14y = 0$
(E) $4x - 5y = 0$
Any hints?
My first attempt was related the tranformation the each coordanaties from line r: 2x-3y = 0 with matrix T.
r as a vector: $\vec{r} = \left[\begin{matrix} 2\\-3\end{matrix}\right]$
$\vec{z} = T . \vec{r} =  \left[\begin{matrix} 4\\-5\end{matrix}\right]$ (wrong)
 A: The simplest way to obtain the image of the given line 
$$2x-3y = 0$$
is to obtain images of its $2$ points - why don't select the natural points $A=[0,0],\ B=[3,2]$?
Computing their images we obtain
$$\left( \begin{matrix}
    5 & 2 \\
    4 & 1 \\
    \end{matrix}\right )\cdot
\left( \begin{matrix}
    0 \\
    0 \\
    \end{matrix}\right ) = \color{red}{
\left( \begin{matrix}
    0\\
    0 \\
    \end{matrix}\right )},\quad\left( \begin{matrix}
    5 & 2 \\
    4 & 1 \\
    \end{matrix}\right )\cdot
\left( \begin{matrix}
    3 \\
    2 \\
    \end{matrix}\right ) = \color{red}{
\left( \begin{matrix}
    19\\
    14 \\
    \end{matrix}\right )},$$
Now we write the equation of the line passing through resulting points $\color{red}{(0, 0)}$ and $\color{red}{(19, 14)}$, obtaining the correct result
$$\color{red}{14x-19y=0},$$
which is similar to your incorrect result (B).

A: The line $2x-3y=0$ defines a subspace. The vector $r$ that you used is orthogonal to this line. 
The correct point of view is to look at the line as set of all $(x,y)\in\mathbb R^2$ such that $(x,y)=(x, \frac{2}{3}x)$ since $y=\frac{2}{3}x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing
$\begin{pmatrix} 5& 2\\ 4& 1\end{pmatrix} \begin{bmatrix} 3\\ 2\end{bmatrix}$
should give the answer.
A: It might help to think of the problem this way: If you’ve been given some equations that describe the relationship between the pairs of variables $(x,y)$ and $(x',y')$, how might you express the equation $2x-3y=0$ in terms of $x'$ and $y'$? One way is to solve those first equations for $x$ and $y$, then substitute into $2x-3y$ and simplify. In this case, you’re given that $$\begin{bmatrix}x'\\y'\end{bmatrix} = T \begin{bmatrix}x\\y\end{bmatrix},\tag1$$ so solving for $x$ and $y$ is a simple matter of computing $T^{-1}$.  
You could also work directly with vectors and matrices. The equation of the line can be written as $$\begin{bmatrix}2&-3\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = 0\tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$\begin{bmatrix}2&-3\end{bmatrix}T^{-1}\begin{bmatrix}x'\\y'\end{bmatrix} = 0\tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$\begin{bmatrix}2&-3\end{bmatrix} \begin{bmatrix}5&2\\4&1\end{bmatrix}^{-1}.$$ I trust that you’re able to compute the value of this expression at this stage of your studies.  
Incidentally, the vector $[2,-3]^T$ is orthogonal (normal) to the line. Equation (3) shows that such normal vectors transform differently than do vectors that represent points: If the point transformation is given by $\mathbf v' = M\mathbf v$, then a normal vector $\mathbf n$ transforms as $\mathbf n' = M^{-T}\mathbf n$. (The $-T$ superscript means “inverse transpose.”)
A: STEP 1. Realize that $\{《x, y》: 2x + 3y = 0\}$ is a subspace.
STEP 2. find the basis for the subspace. It is $(1, 2/3)$.
STEP 3. Do matrix multiplication.
$$\left[ \begin{matrix}
    5 & 2 \\
    4 & 1 \\
    \end{matrix}\right ]
\left[ \begin{matrix}
    1 \\
    2/3 \\
    \end{matrix}\right ] =
\left[ \begin{matrix}
    19/3\\
    14/3 \\
    \end{matrix}\right ]$$
STEP 4. Translate back into equation form.
$$P_1 = \alpha \left[ \begin{matrix}
    19/3 \\
    4/3 \\
    \end{matrix}\right ]$$ and $$P_2 = \beta \left[ \begin{matrix}
    19/3 \\
    4/3 \\
    \end{matrix}\right ]$$. Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = \frac{y_2-y_1}{x_2 - x_1}x$$.
