Calculating the area of a cross-section of a tetrahedron Ok, I completely revised my question. For those interested about my purpose with this question, see the older versions.
So, I would like to calculate the area of a cross-section of a tetrahedron. The only things I know are the $4$ points, with each their own $(x,y,z)$ coordinates.
Under this tetrahedron is a line, the $x$-axis. The cross-sections should be perpendicular to the $x$-axis. A cross-section of a tetrahedron can either be a triangle or a quadrilateral. Between the projection of the two left-most points it will be a triangle, and also between the projections of the third and fourth point. Between the second and third point it will be a quadrilateral. For illustrative purposes:

My question is, how can I find the area of a cross-section perpendicular to the $x$-axis?
 A: Ok, I more or less solved it --  but I'm not very happy with the solution. I would like to end up with an expression for the area between the first and second points, dependent on $y$ of course. This should have the form $a + by + cy^2$ (in this case I chose to project it on the $y$-axis instead of the $x$-axis). I explain why.
Lets create a tetrahedron (having volume 1) with the points $P_1 = (1,0,2/3)$, $P_2 = (4,1,0)$, $P_3 = (1,2,4)$ and $P_4 = (0,3,1)$. Note that the $y$-position is respectively $0,1,2,3$. The tetrahedron looks like this:

For the cross-sections between $P_1$ en $P_2$, we need points on the lines $P_1 P_2$, $P_1 P_3$ and $P_1 P_4$ (so three points in total). The equations for the relevant parts of these lines are
$$\begin{align}
P_1 P_2 & = (1-y)P_1 + y P_2  & y \in [0,1] \\
P_1 P_3 & = (1-\frac{y}{2})P_1 + \frac{y}{2} P_3  & y \in [0,\frac{1}{2}] \\
P_1 P_4 & = (1-\frac{y}{3})P_1 + \frac{y}{3} P_3  & y \in [0,\frac{1}{3}] \\
\end{align}$$
The yellow cross-section above is for $y=0.8$. The area of the triangle can now be calculated as follows. Define two sides of the yellow triangle (using the blue dots):
$$\begin{align}
Side1 = P_1 P_2 (0.8) - P_1 P_3 (0.8)\\
Side2 = P_1 P_4 (0.8) - P_1 P_2 (0.8)
\end{align}$$
The area now follows from the cross product between these two, i.e. $A = \frac{1}{2} | Side1 \times Side2 |$.
If you apply this procedure for $y = 0$ with small steps until $y = 1$ and plot the area as a function of $y$, you end up with (blue solid line):

Which is actually the same as the first part of the uniform quadratic B-Spline, i.e. $.5 t^2$ (yellow dashed line). Perhaps you know that these B-Splines can be defined as projections of simplices, so they are also called "Simplex Splines". 
New Question: I should be able to derive this $.5 t^2$ somehow, using the steps described above. How should I do this?
