91影视

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.

We can split this quadrilateral into two triangles, $锊�P_4{P}_2{P}_3$and$锊�P_4{P}_1{P}_2$ .
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" $$\begin{gathered} A_1+A_2=\frac{1}{2}\left(\left(x_2-x_4\right)\left(y_3-y_4\right)-\left(x_3-x_4\right)+\left(y_2-y_4\right)-\left(x_1-x_4\right)\left(y_1-y_4\right)\right) \\ {A}_1+A_2=\frac{1}{2}\left(x_1{y}_2-y_1+x_2+x_2{y}_3-y_2{x}_3+x_3{y}_4+x_4{y}_1-y_4{x}_1\right) \\ {A}_1+A_2=\frac{1}{2}\left(det\left[\begin{array}{cc}{\mathrm{x}}_1& {\mathrm{x}}_2\\ {\mathrm{y}}_1& {\mathrm{y}}_2\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_2& {\mathrm{x}}_3\\ {\mathrm{y}}_2& {\mathrm{y}}_3\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_3& {\mathrm{x}}_4\\ {\mathrm{y}}_3& {\mathrm{y}}_4\end{array}\right]+det\left[\begin{array}{cc}{\mathrm{x}}_4& {\mathrm{x}}_1\\ {\mathrm{y}}_4& {\mathrm{y}}_1\end{array}\right]\right) \end{gathered}$$