Show up to sign the area of a triangle in $\mathbb{R^2}$ with vertices $(x_{0}, y_{0}), (x_{1},y_{1}), (x_{2}, y_{2})$ is
$$\frac{1}{2} \det \begin{bmatrix}
x_{1} – x_{0} & x_{2} – x_{0} \\
y_{1} – y_{0} & y_{2} – y_{0}
\end{bmatrix} $$
I'm having trouble conjuring up the right idea to show this. We don't know if the triangle is an equilateral triangle. Since all I could do is compute the lengths of the sides I would still need to figure out the height of the triangle. I was thinking of trying to use Pythagoras, but in order for me to do that I would need to find out which side provides the greatest height. Some guidance on how to proceed would be appreciated. I don't need the solution, just the process to do it and perhaps the reasoning. Figuring out the determinant is straightforward, it is using just the points that is an issue.
Best Answer
You can think of triangle area as sum of 2 trapezoids, then remove the base.
$$Area = \left|{y_1 + y_0 \over 2}(x_1 - x_0) + {y_2 + y_1 \over 2}(x_2 - x_1) - {y_2 + y_0 \over 2}(x_2 - x_0) \right|$$
The middle term can be rewrite as:
$${(y_2+y_1)(x_2-x_0) - (y_2+y_1)(x_1-x_0) \over 2}$$
$$Area = {1\over2}\left|(y_0-y_2)(x_1 - x_0) - (y_0 - y_1)(x_2 - x_0) \right|$$
Flip the sign inside absolute function, we get your ½ Det expression.