Why is the SVM margin equal to $\frac{2}{\|\mathbf{w}\|}$? I am reading the Wikipedia article about Support Vector Machine and I don't understand how they compute the distance between two hyperplanes.
In the article, 

By using geometry, we find the distance between these two hyperplanes
  is $\frac{2}{\|\mathbf{w}\|}$

I don't understand how the find that result.
 
What I tried
I tried setting up an example in two dimensions with an hyperplane having the equation $y = -2x+5$ and separating some points $A(2,0)$, $B(3,0)$ and $C(0,4)$, $D(0,6)$ . 
If I take a vector $\mathbf{w}(-2,-1)$ normal to that hyperplane and compute the margin with $\frac{2}{\|\mathbf{w}\|}$ I get $\frac{2}{\sqrt{5}}$ when in my example the margin is equal to 2 (distance between $C$ and $D$). 
How did they come up with $\frac{2}{\|\mathbf{w}\|}$ ?
 A: Let $\textbf{x}_0$ be a point in the hyperplane $\textbf{wx} - b = -1$, i.e., $\textbf{wx}_0 - b = -1$. To measure the distance between hyperplanes $\textbf{wx}-b=-1$ and $\textbf{wx}-b=1$, we only need to compute the perpendicular distance from $\textbf{x}_0$ to plane $\textbf{wx}-b=1$, denoted as $r$.
Note that $\frac{\textbf{w}}{\|\textbf{w}\|}$ is a unit normal vector of the hyperplane $\textbf{wx}-b=1$. We have
$$
\textbf{w}(\textbf{x}_0 + r\frac{\textbf{w}}{\|\textbf{w}\|}) - b = 1
$$
since $\textbf{x}_0 + r\frac{\textbf{w}}{\|\textbf{w}\|}$ should be a point in hyperplane $\textbf{wx}-b = 1$ according to our definition of $r$.
Expanding this equation, we have
\begin{align*}
& \textbf{wx}_0 + r\frac{\textbf{w}\textbf{w}}{\|\textbf{w}\|} - b = 1 \\
\implies &\textbf{wx}_0 + r\frac{\|\textbf{w}\|^2}{\|\textbf{w}\|} - b = 1 \\
\implies &\textbf{wx}_0 + r\|\textbf{w}\| - b = 1 \\
\implies &\textbf{wx}_0 - b = 1 - r\|\textbf{w}\| \\ 
\implies &-1 = 1 - r\|\textbf{w}\|\\
\implies & r = \frac{2}{\|\textbf{w}\|}
\end{align*}
A: 
Let $\textbf{x}_+$ be a positive example on one gutter, such that
$$\textbf{w} \cdot \textbf{x}_+ - b = 1$$
Let $\textbf{x}_-$ be a negative example on another gutter, such that $$\textbf{w} \cdot \textbf{x}_- - b = -1$$
The width of margin is the scalar projection of $\textbf{x}_+ - \textbf{x}_-$ on unit normal vector , that is the dot production of $\textbf{x}_+ - \textbf{x}_-$ and $\frac{\textbf{w}}{\|\textbf{w}\|}$
\begin{align}
width & = (\textbf{x}_+ - \textbf{x}_-) \cdot \frac{\textbf{w}}{\|\textbf{w}\|} \\
      & = \frac {(\textbf{x}_+ - \textbf{x}_-) \cdot {\textbf{w}}}{\|\textbf{w}\|} \\
      & = \frac{\textbf{x}_+ \cdot \textbf{w} \,{\bf -}\, \textbf{x}_-\cdot \textbf{w}}{\|\textbf{w}\|} \\
      & = \frac{1-b-(-1-b)}{\lVert \textbf{w} \rVert} \\
      & = \frac{2}{\|\textbf{w}\|}
\end{align}
The above refers to MIT 6.034 Artificial Intelligence
A: The margin equals the shortest distance between the points of the two hyperplanes. Let $\mathbf{x_1}$ be a point of one hyperplane, and $\mathbf{x}_2$ be a point of the other hyperplane.
We want to find the minimal value of $\lVert \mathbf{x}_1 - \mathbf{x}_2 \rVert$.
Since
\begin{align}
\mathbf{w}\cdot\mathbf{x}_1 - b &= 1,\\
\mathbf{w}\cdot\mathbf{x}_2 - b &= -1,
\end{align}
we have
$$\mathbf{w}\cdot(\mathbf{x}_1 - \mathbf{x}_2) = 2.$$
By the Cauchy-Schwarz inequality, we have
$$\lVert \mathbf{w} \rVert \lVert \mathbf{x}_1 - \mathbf{x}_2 \rVert \geq 2,$$
and therefore
$$\lVert \mathbf{x}_1 - \mathbf{x}_2 \rVert \geq \frac{2}{\lVert \mathbf{w} \rVert },$$
where equality holds when $\mathbf{w}$ and $\mathbf{x}_1-\mathbf{x}_2$ are linearly dependent (which is clearly always possible).
