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I'm having a bit of confusion with a passage I'm reading about Maximal Margin Classifiers in the context of Support Vector Machines, which is making me think I need to go back to grade school. Here's the passage:

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For some reason I can't seem to understand why this is the equation of a line. If I use the basic equation I know, $y=mx+b$, then rename $x$ to $X_{1}$, rename $y$ to $X_{2}$, rename $b$ to $\beta_{0}$, rename $m$ to $\beta_{1}$, then I get $X_{2}=\beta_{1}X_{1}+\beta_{0}$, or rearranged as: $\beta_{0}+\beta_{1}X_{1}-X_{2}=0$. Now flipping the sign I could maybe ignore as being unimportant, but why is there an extra coefficient term as well?

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    $\begingroup$ Because you can divide both sides simply by $\beta_2$ (I assume that the coefficients are real numbers). Equation 9.1 is slightly more general then the equation of a line but can made into one . $\endgroup$ Commented Nov 22, 2017 at 22:52

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You could just multiply the equation $\beta_0+\beta_1X_1-X_2=0$ by another constant to give a "visible" coefficient on the $X_2$ as in equation (9.1).

More important, however, is that $y=mx+b$ is not the general equation of a line, because it will never give you a vertical line $x=c$. The form (9.1) is superior in this respect: it will give you vertical lines (when $\beta_2=0$), horizontal lines (when $\beta_1=0$), and everything in between.

(9.1) has the minor disadvantage that different equations will represent the same line, for example, $$1+2X_1+3X_2=0\quad\hbox{and}\quad 10+20X_1+30X_2=0\ ,$$ but as long as you are aware of this you will find that it rarely causes difficulties.

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  • $\begingroup$ Thanks this answer included everything I needed plus some things I didn't even realized I should be asking :) $\endgroup$
    – Austin
    Commented Nov 22, 2017 at 23:03
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A two-dimensional plane $P$ is spanned by $2$-linearly independent vectors say $X_1,X_2$. Since $\mathbb{R}^2$ is two dimensional, then any point $(u,v)$ is in the image of the matrix $A = [X_1 \mid X_2]$ i.e,

$$ \begin{pmatrix} X_1 & X_2 \end{pmatrix} \begin{pmatrix} \beta_1 \\ \beta_2\end{pmatrix} = \begin{pmatrix} u \\ v\end{pmatrix}$$

It we call $(u,v)^T $ say $C_0$ then the above says,

$$ X_1 \beta_1 + X_2 \beta_2 - C_0 = 0$$

Now set $-C_0 = \beta_0$ then we have,

$$X_1 \beta_1 +X_2 \beta_2 + \beta_0 = 0$$

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  • $\begingroup$ If we call $(u,v)'$ as $C_0$, then the latter is a vector (actually, a matrix). How can $C_0$ be equal to $-\beta_0$, which is a scalar? $\endgroup$ Commented Nov 2, 2021 at 5:00

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