Checking whether a point lies on a wide line segment I know how to define a point on the segment, but I have this piece is a wide interval. The line has a width.
I have $x_1$ $y_1$, $x_2$ $y_2$, width and $x_3$ $y_3$
$x_3$ and $x_4$ what you need to check.
perhaps someone can help, and function in $\Bbb{C}$ #
 A: Trying to understand your question, perhaps this picture might help.

You seem to be asking how to find out whether the point $C$ is inside the thick line $AB$.
You should drop a perpendicular from $C$ to $AB$, meeting at the point $D$.  If the (absolute) length of $CD$ is more than half the width of the thick line then $C$ is outside the thick line (as shown in this particular case).  
If the thick line is in fact a thick segment, then you also have to consider whether $D$ is between $A$ and $B$ (or perhaps slightly beyond one of them, if the thickness extends further).  
A: Assuming @Henry's picture summarizes the question asked, for a given thickness $T$ a necessary and sufficient condition is given by the following inequalities:
$$
0\le\overrightarrow{AC}\cdot\overrightarrow{AB}\le AB^2,\qquad\qquad
AB^2AC^2\le T^2AB^2+\left(\overrightarrow{AC}\cdot\overrightarrow{AB}\right)^2.
$$
The first condition ensures that the projection of $C$ on the line $(AB)$ lies between $A$ and $B$ and the second condition ensures that the distance between $C$ and the line $(AB)$ is at most $T$.
To prove this, note that $\overrightarrow{AC}$ must be $\overrightarrow{AC}=u\overrightarrow{AB}+t\overrightarrow{N}$ with $0\le u\le 1$, $t^2\le T^2$ and $\overrightarrow{N}$ a unitary vector orthogonal to $\overrightarrow{AB}$ and try to express the conditions on $u$ and $t$ in terms of $\overrightarrow{AC}$, $\overrightarrow{AB}$, their norms $AC$ and $AB$, and their scalar product $\overrightarrow{AC}\cdot\overrightarrow{AB}$ only.
A: Please check "Line defined by two points" section of 
https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
From there, it is easy to derive the C# function:
double distOfLineDefinedBy2PointsAndPoint(double x1, double y1, double x2, double y2, double x3, double y3) {
        return Math.Abs((y2 - y1) * x3 - (x2 - x1) * y3 + x2 * y1 - y2 * x1) /
                Math.Sqrt(Math.Pow(y2 - y1, 2) + Math.Pow(x2 - x1, 2));
    }

Now, you can use the result of this function and compare it to half of the width.
