In this section we will use direction vectors rather than slope because in $\mathbb{R}^{3} (means \;3d)$ there will also be parallel lines and direction vectors are easy, but slopes are difficult.
Parallel lines have the same direction vector. Given points, $P=\left[\begin{array}{c} 5\\ 3 \end{array}\right]$ , $Q=\left[\begin{array}{c} 2\\ 8 \end{array}\right]$ , express the line as $$\vec{r}=\vec{P}+t\cdot(\vec{Q}-\vec{P}).$$ The slope equivalent is $(\vec{Q}-\vec{P})$ , called the direction vector. Any two lines with the same direction vector are parallel. In this example, $$Q-P=\left[\begin{array}{c} -3\\ 5 \end{array}\right] \Longleftarrow \text{ direction vector}$$ .
Suppose we had chosen to start at point $Q$ instead of point $P$. Then the line would be $\vec{r}=\vec{Q}+t\cdot(\vec{P}-\vec{Q})$. In this case,the direction vector is $\vec{P}-\vec{Q}=\left[\begin{array}{c} 3\\ -5 \end{array}\right].$ It is the same line, but we are walking along it in the opposite direction. As it turns out, if the direction vectors are proportional, then the lines are parallel.
The direction vector of some line could be $\left[\begin{array}{c} 12\\ -20 \end{array}\right]$ and the line would still be parallel to the line above because we can multiply by $-4$ (a proportionality scalar) and arrive back at the original vector. Often a direction vector will be normalized to the unit circle. To normalize it, we make its length=1. That is done by $\text{Unit Normal Vector}=P/\left|P\right|$. For the vector point denoted by $P=\frac{P_x}{P_y}$, then $\frac{P}{\vert P \vert} = \frac{\left(\begin{array}{c}P_x\\P_y \end{array}\right)}{\sqrt{P_x^2+P_y^2}}.$ Remember that you cannot divide with a vector in the denominator, but what we have done here has a scalar in the denominator.