Finding a direction vector Let $u ∈ \mathbb{R}^2$ be the vector $\begin{pmatrix}3\\-4\end{pmatrix}$
. Let $l$ be the line with equation
$u · v = 50.$

Express $l$ in parametric form.

Solution: A direction vector for $l$ is $\begin{pmatrix}4\\3\end{pmatrix}$
. The point $\begin{pmatrix}6\\-8\end{pmatrix}$
lies on $l$. Hence $l$ consists of all points
of the form

$\begin{pmatrix}6\\-8\end{pmatrix}$ $+ λ\begin{pmatrix}4\\3\end{pmatrix}$.

What I want to know is how is it immediately known that $\begin{pmatrix}4\\3\end{pmatrix}$ is a direction vector for $l$? I thought you would have to find two points on $l$, say $x=\begin{pmatrix}6\\-8\end{pmatrix}$ and $y=\begin{pmatrix}18\\1\end{pmatrix}$ and compute $y-x=\begin{pmatrix}3\\9\end{pmatrix}$ Is this still correct?
Your longer way is correct, only notice that $y-x=\begin{pmatrix}12\\9\end{pmatrix}$.

But any vector orthogonal to $u$ will do, because if $x$ and $y$ both satisfy the equation of the line, then:
$$
u\cdot(x-y)= u\cdot x-u\cdot y =0.
$$
And if $u=\begin{pmatrix}a\\b\end{pmatrix}$, an obvious vector orthogonal to it is 
$\begin{pmatrix}-b\\a\end{pmatrix}$.
Let $v = \begin{pmatrix}x\\ y\end{pmatrix}.$ Then, the line $l$ is described as follows:

$$u \cdot v = 50 \Rightarrow x - 4y = 50.$$

Hence, a parametric representation is:

$$\begin{cases}
x = 50 -4 k \\
y = k
\end{cases},
$$

with $k \in \mathbb{R}$.