8. Derive the formula expressing the area of a triangle through the coordinates of its three vertices: $A_{1}\left(x_{1} ; y_{1}\right), A_{2}\left(x_{2} ; y_{2}\right)$, $A_{3}\left(x_{3} ; y_{3}\right)$. (The vertices are numbered in the order of traversal around the triangle's perimeter counterclockwise.)
8. Let's first consider the case when vertex $A_{3}$ coincides with the origin, and points $A_{1}$ and $A_{2}$ are positioned such that $A_{1} \widehat{O X}\beta$, then our formula gives the area of triangle $O A_{1} A_{2}$ with a negative sign.

It is easy to see that the area $S$ of triangle $A_{1} A_{2} A_{3}$ with arbitrarily located vertices will be equal to the algebraic sum of the areas $\quad S_{\triangle O A_{1} A_{2}}+S_{\triangle O A_{2} A_{3}}+$ $+S_{\triangle O A_{3} A_{1}}$, taken with the signs as above (Fig. 37). Therefore,

$S=\frac{1}{2}\left(\left|\begin{array}{ll}x_{2} & y_{2} \\ x_{3} & y_{3}\end{array}\right|+\left|\begin{array}{ll}x_{3} & y_{3} \\ x_{1} & y_{1}\end{array}\right|+\left|\begin{array}{ll}x_{1} & y_{1} \\ x_{2} & y_{2}\end{array}\right|\right)$,

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-092.jpg?height=551&width=620&top_left_y=250&top_left_x=1095)

Fig. 37

or in the form of a third-order determinant:

$$
S=\left|\begin{array}{lll}
1 & x_{1} & y_{1} \\
1 & x_{2} & y_{2} \\
1 & x_{3} & y_{3}
\end{array}\right|
$$