<s>[INST] Given the points $A(2a, a)$ and $B(2b, b)$ find the coordinates of the point $M$ such that $\vec{AM} = 3 \vec{MB}$. Consider the points:
$$A(2a, a) \hspace{2cm} B(2b, b)$$
with $a \ne b$ and $a, b \in \mathbb{R}$. Find the point $M(x, y)$ such that $\vec{AM} = 3 \vec{MB}$.
First thing I tried was to plot everything we know so far. Here's what the graph looks like with $a = 2$ and $b = 4$:

So it's clear that what I have to do is find the point $M(x, y)$ on the segment $[AB]$ such that we have $\vec{AM} = 3 \vec{MB}$. I thought that an interactive graph would be helpful to see how thing move around, so here it is.
What I did first was create the vectors
$$\vec{OA} = \begin{pmatrix}
           2a \\
           a
         \end{pmatrix} 
\hspace{2cm}
\vec{OB} = \begin{pmatrix}
           2b \\
           b
         \end{pmatrix}$$
Then I got the vector $\vec{AB}$:
$$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix}
           2(b - a) \\
           b - a
         \end{pmatrix}$$
Now since we need $\vec{AM} = 3\vec{MB}$ that means that the point $\vec{AM}$ is situated at three quarters length of the whole vector $\vec{AB}$. So since we now have the vector $\vec{AB}$ we just need to shorten this vector, while keeping its direction. So I created the matrix:
$$X = \begin{pmatrix}
           \dfrac{3}{4} & 0 \\
           0 & \dfrac{3}{4}
         \end{pmatrix}$$
The matrix $X$ squishes each vector to three quarters of its length.
So now we have:
$$\vec{AM} = X \cdot \vec{AB} =
\begin{pmatrix}
           \dfrac{3}{4} & 0 \\
           0 & \dfrac{3}{4}
         \end{pmatrix}
\begin{pmatrix}
           2(b - a) \\
           b - a
         \end{pmatrix}
=
\begin{pmatrix}
           \dfrac{3(b - a)}{2} \\
           \dfrac{3(b - a)}{4}
         \end{pmatrix}$$
So we can now find the coordinates of $M$.
$$\vec{OM} = \vec{OA} + \vec{AM} =  \begin{pmatrix}
           2a \\
           a
         \end{pmatrix}
+
\begin{pmatrix}
           \dfrac{3(b - a)}{2} \\
           \dfrac{3(b - a)}{4}
         \end{pmatrix}
=
\begin{pmatrix}
           \dfrac{a+3b}{2} \\
           \dfrac{a+3b}{4}
         \end{pmatrix}$$
So we can conclude that the point $M$ is at:
$$M \bigg ( \dfrac{a+3b}{2}, \dfrac{a+3b}{4} \bigg )$$
My question is: Is this correct? We've never used matrices like this before in class, it's just an idea that came to me, so I have no idea if my work is correct, even though it seems right. And if it is correct, what is another way of solving this problem? I couldn't come up with any other solution, so that's why I had to do this "slight of hand".
 [/INST] What you're doing looks correct. Using vectors and matrices is a valid way to proceed. However, as you requested, here is an alternate method to consider.
The line going through $A$ and $B$ has a slope of
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{b - a}{2b - 2a} = \frac{1}{2} \tag{1}\label{eq1A}$$
Thus, the general equation for it would be
$$y = \left(\frac{1}{2}\right)x + c \tag{2}\label{eq2A}$$
Using $y = a$ and $x = 2a$ gives that $c = 0$, so $y = \left(\frac{1}{2}\right)x \implies x = 2y$. Note this can be parametrized by having $y = a + t$ so $x = 2a + 2t$, giving
$$f(t) = (2a + 2t, a + t) \tag{3}\label{eq3A}$$
with $f(0) = A(2a, a)$ and $f(b - a) = B(2b, b)$. Since $M$ is three-quarters of the way between $A$ and $B$, then the point is
$$\begin{equation}\begin{aligned}
M(x,y) & = f\left(\left(\frac{3}{4}\right)\left(b-a\right)\right) \\
& = \left(2a + 2\left(\frac{3}{4}\right)\left(b-a\right), a + \left(\frac{3}{4}\right)\left(b-a\right)\right) \\
& = \left(\frac{4a}{2} + \frac{3b-3a}{2}, \frac{4a}{4} + \frac{3b-3a}{4}\right) \\
& = \left(\frac{a + 3b}{2}, \frac{a + 3b}{4}\right)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This matches your answer of $M\left(\frac{a + 3b}{2}, \frac{a + 3b}{4}\right)$.
</s>