Coordinate Geometry - Area of a Quadrilateral

Question

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 $69$.

Answer 1

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 $$\frac{1}{2}\sqrt{\left|\vec{AB}\right|^2\left|\vec{AC}\right|^2-\left(\vec{AB}\cdot \vec{AC}\right)^2}=\frac{1}{2}\sqrt{(1+(-7)^2)\cdot((-8)^2+(-5)^2)-27^2}.$$

Answer 2

There's a neat method we can use in this case called the shoelace formula (which is derived from just adding triangles together, and is applicable in the case where we just have list of the coordinate points).

Application of this formula gives us that the area is

$$\begin{align}A&=\frac{1}{2}\left|x_1y_2+x_2y_3+x_3y_4+x_4y_1-x_2y_1-x_3y_2-x_4y_3-x_1y_4\right| \\&=\frac{1}{2}\left|5\cdot(-4)+6\cdot(-2)+(-3)\cdot7+(-4)\cdot3-6\cdot3-(-3)\cdot(-4)-(-4)\cdot(-2)-5\cdot7\right| \\&=\frac{1}{2}\left|-20-12-21-12-18-12-8-35\right| \\&=\frac{1}{2}\left|-138\right| \\&=69 \end{align}$$

Answer 3

From the diagram:

$$11\times 10-\frac12 \times 9 \times 4 - 4 \times 1 - \frac12 \times 1 \times 7 -\frac12 \times 9 \times 2 - 2 \times 1 - \frac12 \times 1 \times 9$$ $$= 110-18-4 -3.5-9-2-4.5 =69$$