The area of the quadrilateral whose vertices are $(2,1)$....

Question

The area of the quadrilateral whose vertices are $(2,1)$, $(-1,3)$, $(-3,-1)$ and $(3,-4)$ is

My attempt:

I guess the area of quadrilateral $$=\dfrac {1}{2} |x_1y_2 - x_2y_1 + x_2y_3 -x_3y_2 +x_3y_4-x_4y_3 +x_4y_1 - x_1y_4|$$.

An I right ?

Answer 1

You may use the shoelace formula, of course. Or you may prove it:

The area of the given quadrilateral is clearly $$ 6\cdot 7-\left(4+3+2+\frac{5}{2}+9\right) = \color{red}{\frac{43}{2}}.$$

You may also apply Pick's theorem: $$ [ABCD] = \color{purple}{19}+\frac{\color{red}{7}}{2}-1.$$