Question

Consider the quadrilateral in the accompanying figure, with vertices ${\mathit{P}}_{\mathbf{i}}\mathbf{=}\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{,}{\mathbf{y}}_{\mathbf{i}}\mathbf{)}$, for $\mathit{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{4}$. Show that the area of this quadrilateral is

$\frac{\mathbf{1}}{\mathbf{2}}\left(det\left[\begin{array}{cc}{x}_1& x_2\\ y_1& y_2\end{array}\right]+det\left[\begin{array}{cc}{x}_2& x_3\\ y_2& y_3\end{array}\right]+det\left[\begin{array}{cc}{x}_3& x_4\\ y_3& y_4\end{array}\right]+det\left[\begin{array}{cc}{x}_4& x_1\\ y_4& y_1\end{array}\right]\right)$.

Short answer

Therefore,

$A_1+A_2=\frac{1}{2}\left(det\left[\begin{array}{cc}{\mathrm{x}}_1& {\mathrm{x}}_2\\ {\mathrm{y}}_1& {\mathrm{y}}_2\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_2& {\mathrm{x}}_3\\ {\mathrm{y}}_2& {\mathrm{y}}_3\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_3& {\mathrm{x}}_4\\ {\mathrm{y}}_3& {\mathrm{y}}_4\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_4& {\mathrm{x}}_1\\ {\mathrm{y}}_4& {\mathrm{y}}_1\end{array}\right]\right)$and the area of quadrilateral is showed.

Step-by-step solution

Definition. 

The area of a Quadrilateral is used to obtain the area of any type of quadrilateral, whether it is a square, trapezium or a triangle.

A quadrilateral constitutes four sides, four angles, and four vertices.

To show that 12det[x1x2y1y2]+det[x2x3y2y3]+det[x3x4y3y4]+det[x4x1y4y1]

We can split this quadrilateral into two triangles, $ﲔP_4{P}_2{P}_3$and$ﲔP_4{P}_1{P}_2$ .
The area of the first triangle is

$$\begin{gathered} A_1:=\frac{1}{2}\left|det\left[p_2-p_4{p}_3-p_4\right]\right| \\ {A}_1:=\frac{1}{2}\left|det\left[\begin{array}{cccc}{x}_2-& x_4& x_3-& x_4\\ y_2-& y_4& y_3-& y_4\end{array}\right]\right| \\ {A}_1:=\frac{1}{2}\left|\left(x_2-x_4\right)\left(y_3-y_4\right)-\left(x_3-x_4\right)\left(y_2-y_4\right)\right| \end{gathered}$$

Similarly, the area of the second triangle is

$$\begin{gathered} A_2:=\frac{1}{2}\left|det\left(p_1-p_4{p}_2-p_4\right)\right| \\ {A}_2:=\frac{1}{2}\left|det\left[\begin{array}{cccc}{x}_1-& x_2& x_2-& x_4\\ y_1-& y_4& y_2-& y_4\end{array}\right]\right| \\ {A}_2:=\frac{1}{2}\left|\left(x_1-x_4\right)\left(y_2-y_4\right)-\left(x_2-x_4\right)\left(y_1-y_4\right)\right| \end{gathered}$$

Both computations were done in regards to positive orientation, so we can leave out both module signs. Thus, the area of the given quadrilateral is

role="math" localid="1659522425308" $$\begin{gathered} A_1+A_2=\frac{1}{2}\left(\left(x_2-x_4\right)\left(y_3-y_4\right)-\left(x_3-x_4\right)+\left(y_2-y_4\right)-\left(x_1-x_4\right)\left(y_1-y_4\right)\right) \\ {A}_1+A_2=\frac{1}{2}\left(x_1{y}_2-y_1+x_2+x_2{y}_3-y_2{x}_3+x_3{y}_4+x_4{y}_1-y_4{x}_1\right) \\ {A}_1+A_2=\frac{1}{2}\left(det\left[\begin{array}{cc}{\mathrm{x}}_1& {\mathrm{x}}_2\\ {\mathrm{y}}_1& {\mathrm{y}}_2\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_2& {\mathrm{x}}_3\\ {\mathrm{y}}_2& {\mathrm{y}}_3\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_3& {\mathrm{x}}_4\\ {\mathrm{y}}_3& {\mathrm{y}}_4\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_4& {\mathrm{x}}_1\\ {\mathrm{y}}_4& {\mathrm{y}}_1\end{array}\right]\right) \end{gathered}$$