Proving this elegant formula for the area of a parallelogram using the equations of its bounding lines My textbook states that the area of a parallelogram bounded by $y=m_1x+c_1$, $y=m_1x+c_2$, $y=m_2x+d_1$, and $y=m_2x+d_2$ is:
$$area=\left|\frac{(c_1-c_2)(d_1-d_2)}{m_1-m_2}\right|$$
I don't understand how to derive this result. The only thing I can do with these lines is take their perpendicular distance, which is useless because the parallelogram is inclined and not rectangular. Plus, I'll be stuck with a $\sqrt{1+m^2}$ in the denominator which I can't find way to get rid of.
Any advice is appreciated. Thanks!
 A: First, observe that a parallelogram with altitudes, say, $p$ and $q$, and angle $\theta$ at some vertex, can be thought of as having (1) "base" $q/\sin\theta$ and altitude $p$, or (2) "base" $p/\sin\theta$ and altitude $q$. 

Either way, we have

$$\text{A parallelogram with altitudes $p$ and $q$, enclosing angle $\theta$, has area $\frac{pq}{\sin\theta}$.}\tag{$\star$}$$


Now, a version of the result follows immediately when the line equations are in "normal" form:

In this form, $r$ is the (signed) distance from the line to the origin, and $\theta$ is the angle of inclination of the line's "normal" (perpendicular) to the $x$-axis. (We allow $r$ to be negative, in which case we think of the normal as pointing the other way through the origin.)
Suppose the sides of our parallelogram are lines with these normal equations:
$$\begin{align}
x \cos \alpha + y \sin\alpha = p_1 \qquad
x \cos \beta + y \sin\beta = q_1 \\
x \cos \alpha  + y \sin\alpha  = p_2 \qquad
x \cos \beta  + y \sin\beta  = q_2
\end{align}$$
Since $p_1$ and $p_2$ are the signed distances from one set of parallel edge-lines to the origin, $|p_1 - p_2|$ is the distance between the lines themselves; that is, it's the corresponding altitude. Likewise, $|q_1-q_2|$ is the other altitude. A little angle chasing should convince you that the angle between two lines is equal to the angle between their normals ... well ... plus-or-minus some multiple of $180^\circ$, to account for wrong-way-pointing normals, or negatively-measured angles, etc. Thus, the adjacent sides of the parallelogram enclose an angle of $|\alpha - \beta \pm k\cdot 180^\circ|$ . Consequently, by $(\star)$,

$$\text{area of parallelogram} = \left|\frac{(p_1-p_2)(q_1-q_2)}{\sin(\alpha-\beta)}\right| \tag{$\star\star$}$$

(Note that taking the sine of the angle, and the absolute value of the result, allows us to ignore any pesky $180^\circ$s.)
To get the formula in the question, we simply need to recast our line equations into slope-intercept form:
$$x \cos\theta + y \sin\theta = r \quad\to\quad y = -\frac{\cos\theta}{\sin\theta}x + \frac{r}{\sin\theta}$$
So, 
$$\begin{align}
c_i = \frac{p_i}{\sin\alpha} &\quad\to\quad c_1 - c_2 = \frac{p_1-p_2}{\sin\alpha} \\[4pt]
d_i = \frac{q_i}{\sin\beta} &\quad\to\quad d_1 - d_2 = \frac{q_1-q_2}{\sin\beta} \\[4pt]
m_1 = -\frac{\cos\alpha}{\sin\alpha} \quad m_2 = -\frac{\cos\beta}{\sin\beta} &\quad\to\quad m_1-m_2 = \frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\sin\alpha\sin\beta} = \frac{\sin(\alpha-\beta)}{\sin\alpha\sin\beta}
\end{align}$$
and the result follows. $\square$
A: Assume that $m_1$ and $m_2$ are distinct, so that none of the lines is parallel to another one and the parallelogram they bound is well-defined.
The intersection between $y=m_1x+c_1$ and $y=m_2x+d_1$ is the following:
$$A=\left(\frac{d_1-c_1}{m_1-m_2},\frac{m_1d_1-c_1m_2}{m_1-m_2}\right).$$
The intersection between $y=m_1x+c_1$ and $y=m_2x+d_2$ is the following:
$$B=\left(\frac{d_2-c_1}{m_1-m_2},\frac{m_1d_2-c_1m_2}{m_1-m_2}\right).$$
The intersection between $y=m_1x+c_2$ and $y=m_2x+d_1$ is the following:
$$C=\left(\frac{d_1-c_2}{m_1-m_2},\frac{m_1d_1-c_2m_2}{m_1-m_2}\right).$$
Therefore, one gets:
$$\overrightarrow{AB}=\left(\frac{d_2-d_1}{m_1-m_2},m_1\frac{d_2-d_1}{m_1-m_2}\right),\overrightarrow{AC}=\left(\frac{c_1-c_2}{m_1-m_2},m_2\frac{c_1-c_2}{m_1-m_2}\right).$$
Notice that the desired parallelogram is spanned by $\overrightarrow{AB}$ and $\overrightarrow{AC}$ so that it has area:
$$\left|\det\left(\overrightarrow{AB},\overrightarrow{AC}\right)\right|=\left|\frac{d_2-d_1}{m_1-m_2}m_2\frac{c_1-c_2}{m_1-m_2}-m_1\frac{d_2-d_1}{m_1-m_2}\frac{c_1-c_2}{m_1-m_2}\right|=\left|\frac{(c_1-c_2)(d_2-d_1)}{m_1-m_2}\right|.$$
Whence the result.
Perhaps not the cleverest approach but works just fine and the computations are easy to conduct.
A: You can start by assuming we have a parallelogram in a "normalized" position, by which I mean it has one side parallel to the x-axis, and it's lower left corner is on the origin.  Hence the parallelogram is defined by $y = mx, y = mx + c, y = d, y = 0$, with $c < 0$.
In this case we just want to show that $A = |\frac{c d}{m}|$.  We can argue geometrically: the parallelogram fits snuggly in the rectangle with corners $(0,0), (0,d), (\frac{d-c}{m}, 0), (\frac{d-c}{m},d)$.  This rectangle contains exactly two excess triangles, each of area $\frac{d^2}{2m}$.  Thus the area of the parallelogram is indeed $\frac{(d-c)d}{m} - \frac{d^2}{m} = |\frac{cd}{m}|$
Now see if you can derive the general formula by considering isometries of the plane.  Given a generic parallelogram, find an invertible transformation (rotation and translation) that takes takes it to a 'normalized' parallelogram.  What are the values $c$, $d$ and $m$ for the new parallelogram in terms of the values $c_1, c_2, d_1, d_2, m_1, m_2$? Since the isometry preserves area, plugging into $A = |\frac{cd}{m}|$ will yield the desired formula. 
A: 
This is more or less intuitional... But easy to understand. So first we pick a set of parallel lines and construct a segment parallel to the y-axis, with two endpoints on the parallel lines. This segment is $BH$ in the picture. Obviously we have $BH=d_2-d_1$.
Then, we construct two another segments parallel to $BH$, through the intersections of these two parallel lines with another line. These two segments are $IC$ and $A_1J$(Sorry for the bad naming, I'm rather new to Geogebra).
Immediately you see that the two triangles in brown are congruent. Thus, the original parallelogram $ACA_1B$ has same area with $ICA_1J$.
The rest are easy, you construct a segment $KA_1$ through $A_1$ parallel to the x-axis. The area $S=|A_1K|\cdot|BH|$. $A_1K$ is easy to find, you will just need to find the x-coordinate of $C$ and $A_1$, which are solutions to:
$m_2x_c+d_2=m_1x_c+c_1$, and
$m_2x_a+d_2=m_1x_a+c_2$.
With easy algebraic manipulation you find
$A_1K=x_a-x_c=\frac{c_1-c_2}{m_2-m_1}$, multiplying this with BH and take the absolute value, you will arrive that the final expression!
