Write line in parametric form Let's say I have the line $3x+4y=2$
I want to write it in parametric form. You solve for either x or y.
Solving for x: $x=\frac{2}{3}+\frac{-4t}{3}$
Solving for y: $y=\frac{2}{4}+\frac{-3x}{4}$
In parametric form they become:
$\begin{bmatrix}2/3 \\ 0\end{bmatrix}+-t\begin{bmatrix} 4/3\\ 1\end{bmatrix}$
$\begin{bmatrix}0 \\ 2/4\end{bmatrix}+-t\begin{bmatrix} 1\\ 3/4\end{bmatrix}$
If I plot them, the solve for y line is the one that is the same as 3x+4y=2. I see examples online of people solving for both x/y interchangeably. Am I misunderstanding something? What about if I want to write a line on its parameter form, does the variable that I solve for matter?
 A: Specifically, when you have an implicit equation, you can solve for either $x$ or $y$ to derive a parametric version (only exception being when either $x$ or $y$ does not appear in the implicit equation, for example in the lines $x-2=0$ and $3y+5=0$).  However parametric representations are not unique, so that a single line may be represented by any number of parametric equations.
In general if a single line has two parametric representation $\mathbb a_i + t \mathbb b_i$ for $i=1,2$ then we must have $\mathbb b_1$ is a non-zero multiple of $\mathbb b_2$ and both $\mathbb a_1$ and $\mathbb a_2$ must lie somewhere on the line, which is to say $\mathbb a_1 + t\mathbb b_1 = \mathbb a_2$ for some value of $t$.
In your example you could choose the parameter $t$ to be equal to $x$ or $y$ or indeed some other quantity such as $t=x+y$ giving
$$
\left( \array{x\\y} \right)=
\left( \array{-2\\2} \right)+
t \left( \array{4\\-3} \right)
$$
So including the two you have derived, we have three representations in parametric form.  Incidentally, you have a sign wrong in your two formulae, which should read
$$
\left( \array{x\\y} \right)=
\left( \array{2/3\\0} \right)+
t \left( \array{-4/3\\1} \right) \quad\text{and}\quad
\left( \array{x\\y} \right)=
\left( \array{0\\2/4} \right)+
t \left( \array{1\\-3/4} \right) .
$$
