How do you find the area of parallelogram from only the sides of a equation

Question

If you have 4 equation of the 4 sides of a parallelogram how do you find its area? lets suppose the equations as

$$A1x+B1y+C1=0 -----(1)$$ $$A1x+B1y+C2=0 -----(2)$$ $$A2x+B2y+D1=0 -----(3)$$ $$A2x+B2y+D2=0 -----(4)$$

My thought process started from

$$Area= b*h$$

we can easily find the height, say between (1) and (2) and take (3) or (4) as base

$$h= \frac{\left(C1-C2\right)}{\sqrt{A^2+B^2}}$$

And as for the base maybe we could find the intersection of (1) and (2) with (3) taking (3) as the base using the distance formula we should be able to find the length of the base and hence the area of the parallelogram

But is there any elegant way? I hope this isn't a repost as there are a similar couple of question but based on vectors (i am alright with vectors being used, but introduce me like a 4-year-old, I have only a basic understanding, that too from physics.)

Answer 1

You have a good start.

To find the base, you can take the perpendicular distance between the other set of sides, and divide by the sine of the corner angle (which you can find by a dot product between the side normals, one of them turned by 90°).

If you keep track of everything symbolically, I believe the square roots will even cancel out at the end.

(Your area formula is wrong, by the way -- the factor of $\frac12$ is for a triangle, not a parallelogram).

Answer 2

We first convert equations of all four lines of parallelogram to the form $y = mx + c$ and say we have,

$y = m_1 x + C_1, y = m_1x + C_2$
$y = m_2 x + D_1, y = m_2x + D_2$

Then area of parallelogram is simply $ \ \displaystyle \small \bigg|\frac{(C_1 - C_2) \cdot (D_1 - D_2)}{m_1-m_2}\bigg|$

(provided $m_1, m_2 \ne \infty$ i.e none of the sides are parallel to y-axis. If two lines are parallel to y-axis, it should simplify to $ \ \displaystyle \small |(C_1 - C_2) \cdot (D_1 - D_2)| \ $).

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Here is how you can get to the above expression (where $m_1, m_2 \ne \infty$).

Take lines with slope $m_1$. We know the $y$ intercept of lines are $C_1$ and $C_2$. So the distance between parallel lines will be,

$d_1 = \small |(C_1 - C_2) \cos \theta_1|$ where $\theta_1$ is angle with x-axis.

Similarly, $d_2 = \small |(D_1 - D_2) \cos \theta_2|$.

The side of parallelogram with slope $m_2$ will then be $\displaystyle \small \frac{d_1}{\sin\theta}$ and side with slope $m_1$ will be $\displaystyle \small \frac{d_2}{\sin\theta}$ where $\theta$ is the angle between two adjacent sides of parallelogram.

We know, $\displaystyle \small \tan \theta_1 = m_1 \implies \cos\theta_1 = \pm \frac{1}{\sqrt{1+m_1^2}}, \cos\theta_2 = \pm \frac{1}{\sqrt{1+m_2^2}}$

We also know, $\displaystyle \small \tan \theta = \bigg|\frac{m_2-m_1}{1+m_1 m_2}\bigg| \implies \sin \theta = \bigg|\frac{m_2-m_1}{\sqrt{(1+m_1^2)(1+m_2^2)}}\bigg|$

So area of parallelogram, $\displaystyle \small A = \frac{d_1}{\sin\theta} \cdot \frac{d_2}{\sin\theta} \cdot \sin \theta = \bigg|\frac{(C_1 - C_2) \cdot (D_1 - D_2)}{m_1-m_2}\bigg|$

Answer 3

You found $DE=h$. Find another altitude $BG=h_1$. Find angle $\widehat {CDA}$. then:

$$A_{ABCD}=\frac{h\times h_1}{sin(\widehat {CDA)}}$$

Answer 4

Maybe give a separate answer for my comment to another answer here.

Denote vertices of our parallelogram with $P_{13}$, $P_{14}$, $P_{23}$, $P_{34}$, so that $P_{ij}$ is the solution of the system consisting of the equations $(i)$ and $(j)$.

Then the area is equal (up to sign) to the determinant of the matrix with rows given by vectors $v_{123}$ and $v_{134}$ where $v_{123}$ has source $P_{13}$ and target $P_{23}$ while $v_{134}$ has source $P_{13}$ and target $P_{14}$.

We then calculate in coordinates \begin{align*} P_{13}&=\left(\frac{B1D1-B2C1}{A1B2-A2B1},\frac{A2C1-A1D1}{A1B2-A2B1}\right)\\ P_{23}&=\left(\frac{B1D1-B2C2}{A1B2-A2B1},\frac{A2C2-A1D1}{A1B2-A2B1}\right)\\ P_{14}&=\left(\frac{B1D2-B2C1}{A1B2-A2B1},\frac{A2C1-A1D2}{A1B2-A2B1}\right) \end{align*} so that \begin{align*} v_{123}=P_{23}-P_{13}&=\left(\frac{B2(C1-C2)}{A1B2-A2B1},-\frac{A2(C1-C2)}{A1B2-A2B1}\right),\\ v_{134}=P_{14}-P_{13}&=\left(-\frac{B1(D1-D2)}{A1B2-A2B1},\frac{A1(D1-D2)}{A1B2-A2B1}\right) \end{align*} and $$ \text{area}=\pm\det\begin{pmatrix} \frac{B2(C1-C2)}{A1B2-A2B1}&-\frac{A2(C1-C2)}{A1B2-A2B1}\\ -\frac{B1(D1-D2)}{A1B2-A2B1}&\frac{A1(D1-D2)}{A1B2-A2B1}\end{pmatrix}=\pm\frac{(C1-C2)(D1-D2)}{A1B2-A2B1}. $$

Answer 5

Here is a demonstration of the Jacobian to find the area of parellogram.

To begin, I take the same equations as @MathLover:

$y = m_1 x + C_1, y = m_1x + C_2$
$y = m_2 x + D_1, y = m_2x + D_2$

And, I transform the equations into a new coordinate systems defined in the following way:

$$ p = y-m_1 x$$ $$ q=y-m_2 x$$

Notice that in this new $(p,q)$ plane, our lines are now parallel to the axes. In this plane, the region required is a rectangle:

$p=C_1,p=C_2$

$q=D_1, q=D_2$

The area of the rectangle bound by the lines are hence given as: $ |(C_1 - C_2)(D_1 - D_2)|$. Now, all we have to do is find the factor by which the areas are scaled in our transformation. This is given by the determinant of the Jacobian:

$$ |J| = \begin{vmatrix} \frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}\end{vmatrix}= \begin{vmatrix} -m_1 & 1 \\-m_2 & 1\end{vmatrix}=m_2 - m_1$$

Now, let's write the statement about ratios explicitly:

$$ \frac{A_{square} }{A_{parellogram} } = |m_2-m_1|$$

Hence,

$$ A_{parellogram} = \frac{|C_1 - C_2| |D_1 - D_2|}{|m_2-m_1|}$$

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Here is how I'd motivate the jacobian idea: In the start we had $\int_R dx dy$, by change of variables, our integral becomes $\int \frac{1}{ |m_2 - m_1|} dp dq$. Since the bounds are a rectangle , the computation is easy.