Area Of Polygon Using Coordinates

The shoelace formula, also known as the shoelace algorithm, Gauss'due south area formula, and the surveyor'due south formula, is all that's required to calculate the area of a polygon.

The surface area of a polygon, given the coordinates of its vertices, is given past the formula

$A = \frac{ane}{2} \brainstorm{vmatrix} x_1 & x_2 & x_3 & ... & x_n & x_1 \\ y_1 & y_2 & y_3 & ... & y_n & y_1 \finish{vmatrix},$

where each pair of coordinates from $(x_1, y_1)$ to $(x_n, y_n)$ represents the coordinates of a vertex of a polygon with $n$ vertices.

Upon expansion of the above formula, we get the formula

$A = \frac{i}{two}\Big\lvert(x_1y_2+x_2y_3+\cdots+x_{north-1}y_n+x_ny_1) - (x_2y_1+x_3y_2+\cdots+y_{n-1}x_n+x_1y_n)\Big\rvert.$

It may look like a very difficult formula to think or write down, but in actuality, information technology is very simple to memorize when to add together and when to subtract the coordinate values. Employ the following epitome equally an case:

For all the $\textcolor{#D61F06}{\textsf{red}}$ arrows, multiply the two coordinates connected by the $\textcolor{#D61F06}{\textsf{blood-red}}$ arrows together and so add all other products with the $\textcolor{#D61F06}{\textsf{red}}$ arrow. For the $\textcolor{#3D99F6}{\textsf{blue}}$ arrows, multiply the two coordinates continued by the $\textcolor{#3D99F6}{\textsf{blue}}$ arrows together and and then add together all other products with the $\textcolor{#3D99F6}{\textsf{blue}}$ arrow. Then subtract the sum of the $\textcolor{#3D99F6}{\textsf{bluish}}$ arrows from the sum of the $\textcolor{#D61F06}{\textsf{cherry}}$ arrows and modulus the value. Finally, halve the modulus value to get the area of a polygon with $due north$ sides.

Through this proof, we will demonstrate how the formula is derived from a basic quadrilateral on a Cartesian aeroplane.

Referring to the effigy, let $\mathbf{A}$ be the area of the triangle with the vertices of coordinates $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ . Then, draw a quadrilateral such that the vertices touch the sides of the quadrilateral exactly and that its sides are parallel to the $10$ - and $y$ -axes. Drawing the quadrilateral volition course the triangles $\mathbf{C}$ , $\mathbf{D}$ , and $\mathbf{E}$ . And then, let the area of the quadrilateral be $\mathbf{Q}$ . Then the equation describing the human relationship between the area of $\mathbf{A}$ in terms of $\mathbf{Q}$ , $\mathbf{C}$ , $\mathbf{D}$ , and $\mathbf{E}$ is $\mathbf{A}=\mathbf{Q} - \mathbf{C} - \mathbf{D} - \mathbf{E}.$ Using concepts from the distance formula and the formula for the surface area of a triangle $A=\frac{i}{2}bh$ , we get that $\begin{aligned} \mathbf{R}&=(x_3-x_2)(y_1-y_3)=(x_3y_1+x_2y_3)-(x_3y_3+x_2y_1)\\ -\mathbf{C}&=-\frac{1}{2}(x_3-x_2)(y_2-y_3)=\frac{1}{2}(-x_3y_2-x_2y_3)+\frac{one}{two}(x_3y_3+x_2y_2)\\ -\mathbf{D}&=-\frac{one}{2}(x_3-x_1)(y_1-y_3)=\frac{1}{2}(-x_3y_1-x_1y_3)+\frac{ane}{2}(x_3y_3+x_1y_1)\\ -\mathbf{Due east}&=-\frac{i}{2}(x_1-x_2)(y_1-y_2)=\frac{1}{2}(-x_1y_1-x_2y_2)+\frac{1}{2}(x_1y_2+x_2y_1). \terminate{aligned}$ Collecting the terms and rearranging, information technology gives the states $\mathbf{A}=\frac{1}{2}\big((x_2y_3-x_3y_2)-(x_1y_3-x_3y_1)+(x_1y_2-x_2y_1)\large),$ which can be rewritten as a determinant $\mathbf{A}=\frac{one}{2}\begin{vmatrix} 1 & 1 & one \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{vmatrix}.$ And if the coordinates are written in a clockwise order, the value of the determinant will be $-\mathbf{A}$ .

As expanse tin can never be negative, in club to suit the possibility of a 'negative' area from the determinant, nosotros have to add an accented sign to the formula. So, we then become $\mathbf{A}=\frac{1}{two}\Big\lvert x_1y_2+x_2y_3+x_3y_1-(x_2y_1+x_3y_2+x_1y_3)\Big\rvert,$ which (somewhat) matches the formula $A = \frac{1}{two}\Big\lvert(x_1y_2+x_2y_3+\cdots+x_{n-i}y_n+x_ny_1) - (x_2y_1+x_3y_2+\cdots+y_{due north-1}x_n+x_1y_n)\Large\rvert$ given to a higher place if and but if $n=3$ . $_\square$

This side by side proof is a secondary proof showing that the shoelace formula does non only utilize to triangles only can likewise exist applied to other polygons such every bit quadrilaterals.

The diagram above shows a quadrilateral with the vertices of coordinates $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ , and $(x_4, y_4)$ . The quadrilateral is and so divided into ii triangles by a directly line connecting the points $(x_1, y_1)$ and $(x_3, y_3)$ with the areas of $\mathbf{A}$ and $\mathbf{B}$ . Using the triangle formula we proved above, we get $\begin{aligned} \mathbf{A}&=\frac{1}{two}(x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3)\\ \mathbf{B}&=\frac{i}{2}(x_1y_3+x_3y_4+x_4y_1-x_3y_1-x_4y_3-x_1y_4). \end{aligned}$ As both the areas of the triangles were traced using the coordinates in a counterclockwise direction, both areas are positive and we tin get the area of the quadrilateral $\mathbf{A_{quad}}$ past calculation the 2 areas $\mathbf{A}$ and $\mathbf{B}$ . Equally you may have noticed, the terms $x_3y_1$ and $x_1y_3$ are present in both areas $\mathbf{A}$ and $\mathbf{B}$ and conveniently cancel each other out, giving us the formula $\mathbf{A_{quad}}=\frac{1}{2}(x_1y_2+x_2y_3+x_3y_4+x_4y_1-x_2y_1-x_3y_2-x_4y_3-x_1y_4),$ where the absolute sign can also be practical to avoid 'negative' areas as in $\mathbf{A_{quad}}=\frac{1}{2}\Big\lvert(x_1y_2+x_2y_3+x_3y_4+x_4y_1-x_2y_1-x_3y_2-x_4y_3-x_1y_4)\Big\rvert,$ which matches the formula $A = \frac{1}{2}\Large\lvert(x_1y_2+x_2y_3+\cdots+x_{n-1}y_n+x_ny_1) - (x_2y_1+x_3y_2+\cdots+y_{n-1}x_n+x_1y_n)\Big\rvert$ if and merely if $due north=4$ .

Therefore, we have besides indirectly proven that any polygon can be calculated using the shoelace formula equally any polygon can exist divided into multiple smaller triangles with its three vertices having a coordinate assigned to it. $_\square$

Find the area of the $\Delta ABC$ whose vertices are $A=(0, 3)$ , $B=(-2, 2)$ , and $C=(-v, -three)$ .

Using the formula $A = \frac{one}{2} \begin{vmatrix} x_1 & x_2 & x_3 & ... & x_n & x_1 \\ y_1 & y_2 & y_3 & ... & y_n & y_1 \terminate{vmatrix},$ substitute all the variables in the formula with the known coordinates in the question to get $A=\frac{1}{2} \begin{vmatrix} 0 & -5 & -2 & 0 \\ 3 & -three & 2 & iii \end{vmatrix}.$ Using the shoelace algorithm, we get $\begin{aligned} A &=\frac{1}{2} \Large\lvert(0-ten-vi)-(-15+6+0)\Big\rvert \\ &=\frac{1}{ii} \lvert -16+9 \rvert \\ &=\frac{i}{ii} \lvert -7 \rvert \\ &=\frac{one}{2}(7) \\ &=3.5. \end{aligned}$ And so, the surface area of the triangle is three.5. $_\square$

There are 3 points $A$ , $B$ , and $C$ which are all collinear. Point $A$ has coordinates $(-iv, -v)$ , bespeak $B$ has coordinates $(p, 3p+1)$ , and point $C$ has coordinates $(viii, 19)$ . Find the value of $p$ .