Geometric/Visual Solution - Shortest Vector for which Dot Product = x + 2y = 5. (Strang P21 1.2.26) Much as I referenced this same exercise, I'm questing after an exclusively geometric solution.

Question: If $\color{#0070FF}{\vec{v} = (1,2)}$ draw all vectors $\vec{w} = (x,y)$ in the plane  $x,y$ with $\color{#0070FF}{\vec{v}}\cdot\vec{w} = \color{#228B22}{x + 2y = 5}$. Which is the shortest $\vec{w}$?
Algebraic solution: It's given that $\mathbf{v \cdot w} = \color{#228B22}{x + 2y = 5}$. Thus, $\mathbf{v}$ forsooth lies on $x + 2y = 5$. We need an upper bound on $|\mathbf{w}|$, thus apply the Cauchy-Schwarz Inequality. $  \mathbf{v \cdot w} \leq |\mathbf{v}| \mid\mathbf{w}| \iff \cfrac{\mathbf{v \cdot w}}{|\mathbf{v}|} \leq |\mathbf{w}| \tag{CS$\neq$}$
Thus, $min \;\mathbf{w} \iff  $ (CS$\neq$) becomes an equality $\iff$ $\mathbf{v}, \mathbf{w}$ are collinear $\iff k\mathbf{v} = \mathbf{w}$.

$\large{1.}$ How and why is $k = 1$?
$\large{2.}$ How can I solve this question with but the following picture?
Since $\mathbf{v \cdot w} := 5,$ thus $\mathbf{v \cdot w} > 0 \iff \cos(\text{angle between $\mathbf{v}$ & $\mathbf{w}$})>0$
$ \iff \text{angle between $\mathbf{v}$ & $\mathbf{w}$} \in (0, 90^{\circ}).  $ So $\mathbf{w}$ must be right of/above $\color{magenta}{x + 2y = 0}$.


To answer your first question:
$$
\mathbf{v \cdot w} = 5 \Longrightarrow 
\mathbf{v} \cdot (k\mathbf{v}) = 5 \Longrightarrow 
k \|\mathbf{v}\|^2 = 5 \Longrightarrow k=1
$$
Regarding the second question ...
By definition, we have $\mathbf{v \cdot w} = \|\mathbf{v}\| \cdot \|\mathbf{w}\| \cos\theta$, where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$. 
But we know that $\mathbf{v \cdot w} = 5$, and $\|\mathbf{v}\| = \sqrt 5$, so
$$
\|\mathbf{w}\|\, \cos\theta = 5/{\sqrt 5} = \sqrt 5.
$$
Now  $\|\mathbf{w}\| \cos\theta$ is the length of the projection of $\mathbf{w}$ onto $\mathbf{v}$. So, we are looking for the shortest possible vector $\mathbf{w}$ whose projected length along $\mathbf{v}$ is $\sqrt 5$. It's obvious (I think) that the shortest vector giving the desired projected length must be parallel to $\mathbf{v}$. 
In other words, if the length of $\mathbf{w}$ is to be minimized, $\mathbf{w}$ must be a scalar multiple of $\mathbf{v}$, say $\mathbf{w} = k\mathbf{v}$. $\color{green}{\bigstar}$ But $\|\mathbf{v}\| = \sqrt 5$, so, to get a projected length of $\sqrt 5$, the scalar multiplier $k$ must be 1. So $\mathbf{w} = \mathbf{v}$.
Alternative methodology after $\color{green}{\bigstar}$:  But  $\mathbf{w} = k\mathbf{v} \Longrightarrow \|\mathbf{w}\|\ = k\|\mathbf{v}\|$,
where $\|\mathbf{w}\|\ \geq 0 \Longrightarrow LHS \geq 0 $ as well. So for the shortest $\mathbf{w}$ such that $\mathbf{v \cdot w}  = 5$ , we must choose $k = 1$. Finally, because we already determined in the anterior paragraph $\mathbf{w} = k\mathbf{v},$ thus $ \mathbf{w} = \mathbf{v}.$
