triangle

Use the cross product formula for triangle area:

$\vec{P_1} = (1,3) \\ \vec{P_2} = (6,8) \\ \vec{P_3} = (9,1)$

Use Point 1 as the reference, and calculate delta vectors:

$D_{12} = (5,5) \\ D_{13} = (8,-2)$

The area is half the magnitude of the cross product of the two delta vectors:

$A = \frac{1}{2} |D_{12} \times D_{13}| = \frac{1}{2} 50 = 25$

So the vertices are $(1,3),(6,8)$ and $(9,1)$ .

We can use the shoe lace formula,

$A=\dfrac{1}{2} \begin{vmatrix} x_1y_2~+ x_2y_3~+ x_3y_1~-x_2y_1~-x_3y_2-x_1y_3 \end{vmatrix} =\dfrac{1}{2} \begin{vmatrix} 1(8)~+ 6(1)~+ 9(3)~-6(3)~-9(8)-1(1) \end{vmatrix}=\boxed{25}$

Another method here is to compute for the side lengths using the distance formula , then use the heron's formula or cosine rule or any other method to compute for the area of the triangle.

Using the matrix method (determinant, first two columns are x and y pairs, the last column is 1's)

$A=\frac{1}{2} \left| \begin{array}{c}[lcr] &1&3&1\\6&8&1\\9&1&1 \end{array} \right| = \frac{1}{2} \cdot 50 = 25$