Chapter 11.3, Problem 62AYU

Show that the formula for problem $61$ yields the same result as the formula using problem $59$.

59. Geometry: Area of triangle A triangle has vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$.

The area of triangle is given by the absolute value of $D$, where $D=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& x_2& x_3\\ y_1& y_2& y_3\\ 1& 1& 1\end{array}\right|$. Use this formula to find the area of a triangle with vertices $(2,3)$, $(5,2)$ and $(6,5)$.

61. Geometry: Area of polygon Another approach of finding the area of polygon by using determinant is to use formula

$A=\frac{1}{2}\left(\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|+\left|\begin{array}{cc}{x}_2& y_2\\ x_3& y_3\end{array}\right|+\left|\begin{array}{cc}{x}_3& y_3\\ x_4& y_4\end{array}\right|+\mathrm{...}+\left|\begin{array}{cc}{x}_n& y_n\\ x_1& y_1\end{array}\right|\right)$

Where $(x_1,y_1)$, $(x_2,y_2)$, $\mathrm{....}$, $(x_n,y_n)$ are the $n$ corner points in counter clockwise order. Use this formula to compute the area pf the polygon from the problem $60$ again. Which method do you prefer?