Finding the parametric equation of a line in linear algebra

I will write: a parametric equation of a straight line $r$ has the form:

$$begin{pmatrix}x \ yend{pmatrix}=begin{pmatrix}p_1 \ p_2end{pmatrix}+tbegin{pmatrix}v_1 \ v_2end{pmatrix}=begin{pmatrix}p_1+tv_1 \ p_2+tv_2end{pmatrix}, tinBbb R$$

where $mathbf p=begin{pmatrix}p_1 \ p_2end{pmatrix}in r$ and $mathbf v=begin{pmatrix}v_1 \ v_2end{pmatrix}neq mathbf 0$ a vector parallel to the straight line.

If

$$vec a= begin{pmatrix}-1 \ 2end{pmatrix}, quad vec b= begin{pmatrix}-8 \ 4end{pmatrix}$$

we find for example the straight line

$$mathbf{x}=begin{pmatrix}-1 \ 2end{pmatrix}+tbegin{pmatrix}-8 \ 4end{pmatrix}$$

$l:Bbb R to Bbb R^2$ by $l(t)=a+bt$. Based on $a$ and take the direction $b$.

Note that $b$ is an vector, which has a direction. Intuitively, you can think the line $l$ as the track of a point moving from $(-1,2)$ along the direction of $(-8,4)$.

Another way is more algebra, for any point $(x,y)$ in l, it can be expressed by $frac{y-2}{x-(-1))}=frac{-8}{4}$, then you can find $(x,y)=(x,frac{-8}{4}(x-(-1))+2)=a+bt$ for some $tin Bbb R$

Of course, this is just some explanation, not a proof, since the line passing through $a$ along $b$ is not formally defined. (you can take $a+bt$ as the formal definition)