Comment by degamad
You're assuming that we know that the length of vector (a, -b) is a²+b². We don't know that.
We start by assuming that the position vector (a, -b) has length c. This implies that we can rotate that vector until it becomes the position vector (c, 0).
As you note, we can create the two vectors above from (1, 0) using linear transformation matrices [(a, b), (-b, a)] and [(c, 0), (0, c)]
So we could create the position vector (c, 0) by starting at (1, 0), applying the linear transformation [(a, b), (-b, a)], then applying a rotation to bring it back to the e1 axis.
Thus for some rotation matrix R,
R × [(a, b), (-b, a)] = [(c, 0), (0, c)]
The determinant of a rotation matrix is 1, so the determinant of the left side is 1×(a²+b²), while the determinant of the right side is c², which is how we end up with a²+b²=c².
Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already.
> Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already
You can define the determinant that way. Now the question is why the cross multiplication formula for determinant accurately computes the area.
You can prove that via decomposition into right triangles https://youtu.be/_OiMiQGKvvc?si=TyEge1_0W4rb648b
Or you can go in reverse from the coordinate formula, to prove that the area is correctly predicted by the determinant.