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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.)

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    $\begingroup$ It's not elegant, but I would probably find the four vertices, and compute the area using the "Shoelace Theorem" (or the scalar product) if you have learnt it before. If you want to stay away from vectors you can also use Heron's Formula for one of the triangles formed when you join the diagonals, but it also won't be elegant. $\endgroup$
    – dodoturkoz
    Apr 5 at 12:42
  • $\begingroup$ @dodoturkoz yes i heard of scalar products, and yes I have heard of shoelace theorem thou i never knew the name... ill make sure to get it glued in my brain. But do you need the vertices for the scalar product? $\endgroup$ Apr 5 at 12:51
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    $\begingroup$ I meant the cross product, sorry for the typo. The area of a parallelogram is given by $A_\text{parallelogram}= \left|\vec{a}\times\vec{b}\right|$. You should first convert the vertices in $\mathbb{R}^2$ to $\mathbb{R}^3$. Then, obtain two vectors to compute cross product to calculate the area as follows: $$\left| \begin{pmatrix}a\\b\\0\end{pmatrix} \times \begin{pmatrix}c\\d\\0\end{pmatrix} \right|$$ $\endgroup$
    – dodoturkoz
    Apr 5 at 13:50
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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).

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  • $\begingroup$ Ah now it all makes sense $\endgroup$ Apr 5 at 14:46
  • $\begingroup$ (Also, i think it should be cross product to get the Sin thetha value, but I guess that puts us at equal terms lol) $\endgroup$ Apr 5 at 14:49
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    $\begingroup$ @AdilMohammed: In 2D the cross product is the same as "dot product with one of the operands turned by 90°". $\endgroup$ Apr 5 at 15:25
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enter image description here

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

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

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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)| \ $).


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

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    $\begingroup$ In terms of original coefficients:$$\frac{(C1-C2)(D1-D2)}{A1B2-A2B1}$$ $\endgroup$ Apr 5 at 22:03
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    $\begingroup$ Yes Adil, that is correct. I wrote the equations in $y = mx+c$ form. That makes it really easy to find the area.. $\endgroup$
    – Math Lover
    Apr 6 at 12:13
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    $\begingroup$ so first (1) to find the length you found the difference of intercept then (2) you found the shortest distance by multiplying ${\cos\theta}$ where $thetha$ is the angle the line makes with coordinate axis then (3) to find the length of the side you divided with ${\sin\theta}$ where thetha is angle between the 2 sides of the parallelogram right? $\endgroup$ Apr 6 at 12:28
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    $\begingroup$ Yes the area can be straightaway found using $\big|\frac{(C_1 - C_2) \cdot (D_1 - D_2)}{m_1-m_2}\big|$. The latter part is only to show how one can arrive at that. To arrive at that, yes I used $A = l \cdot b \cdot \sin\theta$. $\endgroup$
    – Math Lover
    Apr 6 at 12:28
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    $\begingroup$ On your last comment, yes that is absolutely correct. $\endgroup$
    – Math Lover
    Apr 6 at 12:30
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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 $$ \text{area}=\pm\det\begin{pmatrix} \frac{B2(C1-C2)}{A1B2-A2B1}&-\frac{A2(C1-C2)}{A1B2-A2B1}\\ -\frac{B1(D1-D2)}{A1B2-A2B1}&\frac{A1(D1-D2)}{A1B2-A2B1}\end{pmatrix}=\pm\frac{(C1-C2)(D1-D2)}{A1B2-A2B1}. $$

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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,

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


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.

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