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I want to show that the solution set of a vector scalar equation (v=tw+z) is a straight line no matter which vectors w and z are. I am also assuming these vectors are in $\mathbb{R}^2$.

I'm not sure where to begin this proof. Would I use $y=mx+b$?

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    You need $w\ne(0,0)$, otherwise it's not true. – 2017-01-24
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    First, do it for the special case when ${\bf z}$ is the zero vector, and ${\bf w}$ is nonzero, but something simple, like ${\bf w} = (1, 2)$. Plot ${\bf v}$ for a few values of $t$. You'll see what to do next. – 2017-01-24
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    Yes, I think $y=mx+b$ is a good place to start. If $\mathbf{w} = \left$, $\mathbf{z} = \left$, and $\mathbf{v} =\left$, can you show that $y=mx+b$ for some $m$ and $b$? There will be one exception. – 2017-01-24
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    I apologize but I'm feeling very clueless. If I do as @avs has suggested, I would get v to be similar to a linear function where w is only affected by the scalar t. But isn't this only is z is the zero vector? (Sorry again, I'm just struggling with this seemingly easy concept. I think I may be overthinking it) – 2017-01-24
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    Yes, it is. But then, when you continue to the case when ${\bf z}$ is nonzero, you will see that all it does is translate the line you formed earlier. The most difficult part of the problem is to understand the parameterization by $t$. Afterwards, the translation by ${\bf z}$ is easy. That's why I suggested so as to separate these two tasks. – 2017-01-24

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If your definition of a line in $\mathbb{R}^2$ is any set of the form $$ L_{m,b}:=\{(x,y)\in\mathbb{R}^2:y=mx+b\} $$ or $$ V_c:=\{(x,y)\in\mathbb{R}^2:x=c\} $$ (vertical lines) then you can proceed like this.

If ${\bf v}=(v_1,v_2)$, ${\bf w}=(w_1,w_2)$ and ${\bf z}=(z_1,z_2)$, then your vector equation is equivalent to \begin{align} v_1=tw_1+z_1\tag{1}\\ v_2=tw_2+z_2\tag{2} \end{align}

Case 1: If $w_1\neq0$, then $(1)$ is equivalent to $$ t=\frac{v_1-z_1}{w_1} $$ and substituting in $(2)$ and rearranging gives $$ v_2=\left(\frac{w_2}{w_1}\right)v_1+\left(z_2-\frac{z_1w_2}{w_1}\right) $$ In this case, the solution set is $L_{m,b}$ where $m=\frac{w_2}{w_1}$ and $b=z_2-\frac{z_1w_2}{w_1}$.

Case 2: If $w_1=0$, then $v_1=z_1$. If $w_2=0$ also, then $v_2=z_2$ is constant so the solution set is the point $\{{\bf z}\}$ which is not a line. If $w_2\neq0$ then the solution set is $V_c$ with $c=z_1$ (since $v_2$ varies through $\mathbb{R}$ when you let $t$ vary through $\mathbb{R}$).

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One of the many definitions of vector space is that vectors are closed under addition, meaning that if $u, v$ are vectors in $\mathbb{R}^2$, then $u+v \in \mathbb{R}^2$. There's another property that deals with scalar multiplication. How can you use these two properties of vector space to prove your result?