We are given some arbitrary solution $(x_0, y_0)$ to $ax-by=1$ and another point $(b,a)$. We are asked to prove that the area of the triangle with vertices at $(0,0)$, $(b,a)$, and $(x_0, y_0)$ is $\frac{1}{2}$. I figured that I should use the equation for a right triangle, and apply some substitution as follows:
$\frac{1}{2}\sqrt{x_0^2+y_0^2}\sqrt{(b-x_0)^2+(a-y_0)^2}$
plugging in $x_0$ and $y_0$ in terms of the diophantine equation doesn't seem to cancel out any terms and the algebra is messy.
The point $(x_0,y_0)$ lies on a line parallel to the opposite edge of the triangle. Therefore you can choose any point on this line and the area will always be the same!
So just choose the point $(\frac{1}{a},0)$ and the triangle then has base $\frac{1}{a}$ and area $a$. It therefore has area $\frac{1}{2}\times a \times \frac{1}{a}=\frac{1}{2}$.