$$\frac 12\left|\begin{matrix} -\frac{(\sqrt 2+1)}{2} &\frac{1-\sqrt 2}{2} & 1 \\ 0 & 0 & 1 \\ -1 & 0 & 1 \end{matrix}\right| $$
With this we calculate the area of a triangle that has vertices:
$$x=-\frac{(\sqrt 2+1)}{2} $$ and $$y=\frac{1-\sqrt 2}{2}$$ Is the point of the first vertex
The other vertices of the triangle are the $ x$-intercepts of the lines, namely $(0,0)$ and $(-1,0)$. So the area is simply
$$\frac 12\left|\begin{matrix} -\frac{(\sqrt 2+1)}{2} &\frac{1-\sqrt 2}{2} & 1 \\ 0 & 0 & 1 \\ -1 & 0 & 1 \end{matrix}\right| $$
So my question is, how does this work to calculate this. I already have experience with determinants but never to calculate surface.
In two dimensions it's more simple: Let $\vec a=(a_x,a_y)$ and $\vec b=(b_x,b_y)$ then the area of the parallelogram spanned by $\vec a$ and $\vec b$ is just the absolute value of the determinant of these vectors, see https://de.wikipedia.org/wiki/Parallelogramm#Beweis_der_Flächenformel_für_ein_Parallelogramm.