I was doing this problem about affine geometry:
Given three points $A, B,C$ no aligned in a affine plane $A$ in $k$, prove that exist a bijective affinity $f:A\longrightarrow{V}$ such $f(A)=(1,0,0), f(B)=(0,1,0)$ and $f(C)=(0,0,1)$, where $V\subset k^3$ is the plane $x+y+z=1$.
I know that, in a affine transformation, exist only one linear application which $f(P+w)=f(P)+m(w)$ where $w$ is a vector. But no clue how to continue. Trying to consider two points and construct any linear application? What I've considered is a linear application $g$ which sent vectors $AB$ and $AC$ to vectors $u, v \in V$, I think this $g$ exist and is only one. Then is it correct to define $g=m$ and $f(X)=Q+m(AX)$?
The plane $A$ can be described parametrically by
$ a(t, s) = A + t (B - A) + s (C - A) $
where $t, s \in \mathbb{R}$
Now define the transformtion $ f (x) = T x $ where $x \in \mathbb{R}^3$
Then the image of the plane points is
$\begin{equation}\begin{split} b(t, s) &= f(a(t,s)) = f ( A + t (B- A) + s(C - A) )\\ &= f(A) + t (f(B)- f(A)) + s(f(C) - f(A)) \\ &= (1 - t - s) f(A) + t f(B) + s f(C) \\ &= \begin{bmatrix} 1 -t - s \\ t \\ s \end{bmatrix} \end{split} \end{equation}$
Note that the plane $B$ is described by $x + y + z = 1 $ because $(1 - t - s) + t + s = 1 $
Finally, substituting,
$T \begin{bmatrix} A, B, C \end{bmatrix} = I $
it follows that $T = \begin{bmatrix} A, B, C \end{bmatrix}^{-1}$