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so that $H_0$ is equidistant from $H_1$ and $H_2$.
However, here the variable $\delta$ is not necessary. So we can set $\delta=1$ to simplify the problem.
$$w\cdot x+b=1 $$ and
$$w\cdot x+b=−1$$
Why is this assumption is taken? If it is taken, we can get the distance between two planes as $2$ directly because both are parallel and differ by $2$. how it is $\frac{2}{\left\|w\right\|}$ instead.
got equations from this https://www.svm-tutorial.com/2015/06/svm-understanding-math-part-3/
After we have
$$w^Tx + b = \pm \delta$$
We can always divide everything by $\delta$,
$$\left( \frac{w}{\delta}\right)^Tx + \left( \frac{b}{\delta}\right)=\pm1$$
Now, we can set $\tilde{w}=\frac{w}{\delta}$ and $\tilde{b}=\frac{b}{\delta}$.
$$\tilde{w}^Tx+\tilde{b}=\pm1$$
This is as if we have set $\delta=1$ from the beginning.
The derivation of the distance formula has been given in equation $(19)$ in the article that you linked to and you might like to be more specific if you can't understand it. The distance should be $\frac{2\delta}{\|w\|}$ if $\delta$ is not set to be $1$.
