Coordinate Geometry - Area of a Quadrilateral What is the area in square units, of a quadrilateral whose vertices are $$(5,3), (6,-4), (-3,-2), (-4,7)?$$

I have tried creating the triangles, but didn't know how to find the diagonal. I wanted to try the shoelace method but I thought it only worked for triangles. The answer that was provided is $69$.
Let $A(5,3), B(6,-4), C(-3,-2), D(-4,7)$.

The area of $\triangle ABC$ is given by
$$\frac{1}{2}\sqrt{\left|\vec{AB}\right|^2\left|\vec{AC}\right|^2-\left(\vec{AB}\cdot \vec{AC}\right)^2}.$$
One can get the area of $\triangle{ACD}$ in the same way as above.

Added : Since $$\vec{AB}=(6-5,-4-3)=(1,-7)$$$$\vec{AC}=(-3-5,-2-3)=(-8-5)$$$$\vec{AB}\cdot \vec{AC}=1\cdot(-8)+(-7)\cdot (-5)=27,$$one has
$$\frac{1}{2}\sqrt{\left|\vec{AB}\right|^2\left|\vec{AC}\right|^2-\left(\vec{AB}\cdot \vec{AC}\right)^2}=\frac{1}{2}\sqrt{(1+(-7)^2)\cdot((-8)^2+(-5)^2)-27^2}.$$
A(x1,y1), B(x2,y2), C(x3,y3), D(x4,y4) are verticies of a quadrilateral either convex or concave (one of the internal angle greater than 180 degrees) taken in order,  then we can use the following elegant formula for calculating it's area. 

A= 1/2*abs((xi-x3) *(y2-y4) -(x2-x4) *(y1-y3))