Take the general form a line: $$ax+by+c=0$$ Now if $c=0$, we can view this quite intuitively as the set of vectors $\langle x,y \rangle$ such $$\langle x,y \rangle\cdot\langle a,b\rangle =0$$ And hence, all the vectors that are perpendicular to $\langle a,b\rangle$.
But what about $c$? How can we incorporate it into this intuitive explanation?
I understand that increasing $c$ will cause the line to move along the vector $\langle a,b\rangle$ (away from the point $(a,b)$), but what is a good way to understand this.
The two starting points I've come up with are:
- We can think of it instead as the set of all vectors $\langle x,y,0 \rangle$ that are orthogonal to $\langle a,b,c\rangle$, though, without seeing this visualised I'm having trouble picturing this.
- We can think of it as the scalar product of $\langle a,b \rangle$ and $\langle x,y\rangle$ being equal to $-c$. So far this still isn't really helping me though. It isn't really easy to visualise.
We have $$ \eqalign{ & \Delta {\bf P} \cdot {\bf n} = 0\quad \Rightarrow \cr & \Rightarrow \quad \left( {{\bf P} - {\bf P}_{\,0} } \right) \cdot {\bf n} = 0\quad \Rightarrow \cr & \Rightarrow \quad {\bf P} \cdot {\bf n} = {\bf P}_{\,0} \cdot {\bf n}\quad \Rightarrow \cr & \Rightarrow \quad ax + by = c \cr} $$ so that $$ c = {\bf P}_{\,0} \cdot {\bf n} $$ is a sort of measure of the (signed) distance of the line from the origin.
If $b \ne 0$, we can divide through by it $$ y = - {a \over b}x + {c \over b} $$ and get that $c$ is proportional to the $y$ intercept.
otherwise we can divide by $a$ and "measure" the distance by the $x$ intercept.
And if you are in the Euclidean plane, then we can put $$ {a \over {\sqrt {a^{\,2} + b^{\,2} } }}x + {b \over {\sqrt {a^{\,2} + b^{\,2} } }}y = {c \over {\sqrt {a^{\,2} + b^{\,2} } }} $$ and
$$ {c \over {\sqrt {a^{\,2} + b^{\,2} } }} $$ is actually the distance of the line from the origin.
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Answering to your comment, ${\bf P}_{\,0}$ can be any point on the line, so in particular it can correspond to $H$.
In any case it is $$ {\bf P}_{\,0} \cdot {\bf n} = {\bf H} \cdot {\bf n} = c $$
Note that if you are working in the affine space we cannot talk of "distance" in the same acception as in the Euclidean space.
And in the Euclidean space, to have that $c$ be the distance of the line from the origin, signed along the direction of ${\bf n}$ you shall first normalize ${\bf n}$.