Random Triangle's Area

The area of a triangle in the $xy$ plane with vertices at $(x_1,y_1),$ $(x_2,y_2),$ and $(x_3,y_3)$ is equal to this. $\dfrac{1}{2}\times\text{abs}\left(\text{ }\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|\text{ }\right)$

Plugging in the points, you get this. $\dfrac{1}{2}\times \text{abs}\left(\text{ }\left| \begin{array}{ccc} 3 & -2 & 1 \\ 4 & 1 & 1 \\ -1 & 6 & 1 \end{array} \right|\text{ }\right)=\dfrac{1}{2}\times20=\boxed{10}$

We find that the side lengths of the triangle are the square root of 10, square root of 50, and square root of 80. Plugging these values into Heron's formula we find that the area is 10.

Actually i used a further variation of Hero's Formula after finding the lengths of the sides using distance formula

Area = 1/4 \sqrt{4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}

lets take (3,-2) as (x1,y1) and (4,1) as (x3,y3) and (-1,6) as (x2,y2) ..... then use this equation wich is used with cordinate geomatry --------> 1/2 |x3(y1-y2) +x1 (y2-y3) +x2(y3-y1)|