Let $A,B,C\in \mathbb{R}^2$ be noncollinear points. Then we have that for every point $P\in\mathbb{R}^2$ there exist $\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}$ such that $P=\alpha_1A+\alpha_2 B+\alpha_3 C$, and $\alpha_1+\alpha_2+\alpha_3=1$ (I've already proven that).
Now, write $P_1,P_2,P_3\in\mathbb{R}^2$ in the above coordinates :$$P_i=\alpha_{i1}A+\alpha_{i2}B+\alpha_{i3}C, \qquad i=1,2,3,$$
with
$$\alpha_{i1}+\alpha_{i2}+\alpha_{i3} =1.$$
Then I must prove that:
The points $P_1,P_2,P_3$ are collinear iff $\det(a_{ij})=0$.
For $\Rightarrow$, I supposed that they are collinear, in which case $P_3-P_1$ and $P_2-P_1$ are linearly dependent. I tried to express these given the coordinates, it became too long and then I got lost. But I think there must be a better way to prove it.
So, any idea?
Best Answer
Hint Since (1) collinearity is preserved by affine transformations and (2) one can choose an affine transformation that maps $A, B, C$ as given resp. to, e.g., $(0, 0), (1, 0), (0, 1)$, we may as well assume that $A, B, C$ are those points.
(To see the latter, note that we can apply the translation that maps $A$ to the origin and then apply the linear transformation with matrix $\pmatrix{\bar{B}&\bar{C}}^{-1}$, where $\bar{B}, \bar{C}$ are resp. the images of $B, C$ under that translation, regarded as column vectors.)
Then, for any point $P_i(x_i, y_i)$, solving the given system defining the barycentric coordinates $\alpha_{ij}$ is easy: $$\alpha_{i1} = -x_i - y_i + 1, \qquad \alpha_{i2} = x_i, \qquad \alpha_{i3} = y_i.$$