Finding the parametric equation of a line in linear algebra

Question

I was given the above question to answer, and I do not know how to do it. If t is the parameter, then I suppose we can create a relationship between a and b using t, but I don't see how that is helpful. Any guidance?

Thank you!

Answer 1

$l:\Bbb R \to \Bbb R^2$ 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 $t\in \Bbb R$

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)

Answer 2

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

$$\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}p_1 \\ p_2\end{pmatrix}+t\begin{pmatrix}v_1 \\ v_2\end{pmatrix}=\begin{pmatrix}p_1+tv_1 \\ p_2+tv_2\end{pmatrix}, \ t\in\Bbb R$$

where $\mathbf p=\begin{pmatrix}p_1 \\ p_2\end{pmatrix}\in r$ and $\mathbf v=\begin{pmatrix}v_1 \\ v_2\end{pmatrix}\neq \mathbf 0$ a vector parallel to the straight line.

If

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

we find for example the straight line

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