Question

Find the area of the triangle whose vertices are $(-2,3),(3,2)$ and $(-1,-8)$ by using determinant method.

Answer 1

Area of triangle is
$△=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
where $\left(x_1{y}_1\right)=(-2,-3);\left(x_2{y}_2\right)=(3,2);(x_3,y_3)=(-1,-8)$
$\therefore △=\frac{1}{2}\left|\begin{array}{ccc}-2& -3& 1\\ 3& 2& 1\\ -1& -8& 1\end{array}\right|$
Expanding $△$ using $R_1$ (first row)
$△=\frac{1}{2}[-2(2+8)+3(3+1)+1(-24+2\left)\right]$
$=\frac{1}{25}[-20+12-22]=\frac{1}{2}|-30|=\frac{30}{2}$
$△=15sq.units$.