Mapping a point of a triangle to infinity If I have three points that define a triangle $\{(0,0), (1,0), (0,1)\}$, how can I map one of these points (say $(0,1)$) to infinity while keeping the other two points regular (i.e the determinate at these points isn't $0$).  My current best effort is the scaling matrix with the y-scaling as $1/\varepsilon$ and then taking the limit as epsilon tends to $0$, but in this case my determinate approaches infinity for all points.  Is what I am attempting even possible?
Cheers!
 A: The best way I know to think about this is with projective geometry. So we think of the plane as points $[x,y,1]$ and points at infinity as points $[1,b,0]$ or $[0,1,0]$. A projective transformation that leaves
$[0,0,1]$ and $[1,0,1]$ fixed and maps $[0,1,1]$ to $[0,1,0]$ is given by the nonsingular matrix
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1\end{bmatrix}.$$
That is, $f(x,y,z) = (x,y,-y+z)$, so $f([x,y,1]) = \left[\dfrac x{1-y}, \dfrac y{1-y},1\right]$.
So you want the mapping
$$F(x,y) = \left(\frac x{1-y},\frac y{1-y}\right).$$
As you can check, $F(0,0) = (0,0)$, $F(1,0)=(1,0)$, and $F(0,1)$ is at infinity (indeed, the point at infinity on the $y$-axis).
A: Not sure what your end-goal is or what exactly you mean by mapping, but I think you can map $(0,0)$ to infinity and keep the other points "regular" by the following mapping function:
$$
f_1(x,y)=\frac{1}{x+iy}
$$
So $f_1(0,0)=\infty$, $f_1(1,0) = 1$, and $f_1(0,1)=-i$. So your new points would be $\{(\infty,\infty),(1,0),(0,-1)\}$. Maybe this isn't what you're looking for?
