I'm getting started with 2D graphics. I've read that a straight line can be expressed as the following implicit equation.
$\vec{N} \cdot \vec{L} + d = 0$
Where, $\vec{N}$ is normal vector and $\vec{L}$ is the position vector for a point on the line. I have confusion with $d$. Is it the distance of the line from the origin along the normal vector or against the normal vector? In other words, for a straight line $\{4\hat x,5\hat y\} \cdot \vec{L} + 4 = 0$ what will be the correct graphical representation, left one or the right one?
The greenish blue line is the normal and the red one is the intended straight line drawn with cairo-graphics and GTK+.
Hint:
$\vec L$ is a generic vector $(x,y)^T$. Lock at what becomes your equation if $ d=0$, than do the same with a $d \ne 0$.
If $\vec N=(4,5)^T$ ( as in your example) , than for $d=0$ the equation becomes $ 4x+5y=0$ that, as you noted, is the equation of a line that passes thorough the origin since $(0,0)$ satisfies it.
If $d \ne 0$ the equation becomes $4x+5y+d=0$ ad you can see the for $x=0$ the equation is satisfied if $y=-d/5$ so the line intersect the $y$ axis at the point $(0,-d/5)$. In the same way you can find the intercept of the line with the $x$ axis that is $(-d/4,0)$.
If $\vec N=(a,b)$ is a unitary vector ( i.e. $\sqrt{a^2+b^2}=1$) than $|d|$ is the distance of the line from the origin.
This is the simpler interpretation of $d$.