I have $2$ triangles ($6$ dots) on a $2D$ plane.
The points of the triangles are: a, b, c and x, y, z
I would like to find a matrix, using I can transform every point in the 2D space.
If I transform a, then the result is x. For b the result is y, and for c the result is z
And if there is a given d point, which is halfway from a to b, then after the transformation the result should be between x and y halfway.
I've tried to solve it according to NovaDenizen's solution, But the result is wrong.
The original triangle:
$$
a =
\left[
\begin{array}{ccc}
-3\\
0\\
\end{array}
\right]
$$
$$
b =
\left[
\begin{array}{ccc}
0\\
3\\
\end{array}
\right]
$$
$$
c =
\left[
\begin{array}{ccc}
3\\
0\\
\end{array}
\right]
$$
The x, y, z dots:
$$
x =
\left[
\begin{array}{ccc}
2\\
3\\
\end{array}
\right]
$$
$$
y =
\left[
\begin{array}{ccc}
3\\
2\\
\end{array}
\right]
$$
$$
z =
\left[
\begin{array}{ccc}
4\\
3\\
\end{array}
\right]
$$
I've created a figure:
I tried to transform the (0, 0) point, which is halfway between a and b, but the result was (3, 3.5) instead of (3, 3)
The T matrix is:
$$\left[
\begin{array}{ccc}
1/3 & 1/6 & 0\\
0 & -1/2 & 0\\
3 & 3,5 & 1\\
\end{array}
\right]$$
Best Answer
The transformation you're looking for has this form:
$$\left[ \begin{array}{ccc} t_1 & t_2 & t_3\\ t_4 & t_5 & t_6\\ 0 & 0 & 1\\ \end{array} \right] \left[ \begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \\ 1 & 1 & 1 \\ \end{array} \right] = \left[\begin{array}{ccc} x_1 & y_1 & z_1\\ x_2 & y_2 & z_2\\ 1 & 1 & 1 \end{array} \right] $$
or
$${\bf T A} = {\bf X}$$ so $${\bf T} = {\bf X}{\bf A}^{-1}$$
Now $\bf T$ is a transformation matrix you can use on any point, like $${\bf T}\left[\begin{array}{c}a_1\\a_2\\1\end{array}\right] = \left[\begin{array}{c}x_1\\x_2\\1\end{array}\right]$$
It's linear, so it has the halfway point property you were looking for.
$$ {\bf A} = \left[\begin{array}{ccc} -3 & 0 & 3\\ 0 & 3 & 0\\ 1 & 1 & 1\\ \end{array}\right] $$ $$ \left[\begin{array}{ccc|ccc} -3 & 0 & 3 & 1 & 0 & 0\\ 0 & 3 & 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 0 & 0 & 1\\ \end{array}\right] $$ $$ \left[\begin{array}{ccc|ccc} 1 & 0 & -1 & -\frac13 & 0 & 0\\ 0 & 3 & 0 & 0 & 1 & 0\\ 0 & 1 & 2 & \frac13 & 0 & 1\\ \end{array}\right] $$
$$ \left[\begin{array}{ccc|ccc} 1 & 0 & -1 & -\frac13 & 0 & 0\\ 0 & 1 & 0 & 0 & \frac13 & 0\\ 0 & 0 & 2 & \frac13 & -\frac13 & 1\\ \end{array}\right] $$
$$ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -\frac16 & -\frac16 & \frac12\\ 0 & 1 & 0 & 0 & \frac13 & 0\\ 0 & 0 & 1 & \frac16 & -\frac16 & \frac12\\ \end{array}\right]\\ \text{so } {\bf A}^{-1} = \left[\begin{array}{ccc} -\frac16 & -\frac16 & \frac12\\ 0 & \frac13 & 0\\ \frac16 & -\frac16 & \frac12\\ \end{array}\right]$$
$${\bf XA}^{-1} = \left[\begin{array}{ccc} 2 & 3 & 4\\ 3 & 2 & 3\\ 1 & 1 & 1\\ \end{array}\right] \left[\begin{array}{ccc} -\frac16 & -\frac16 & \frac12\\ 0 & \frac13 & 0\\ \frac16 & -\frac16 & \frac12\\ \end{array}\right] = \left[\begin{array}{ccc} \frac13 & 0 & 3\\ 0 & -\frac13 & 3\\ 0 & 0 & 1\\ \end{array}\right] $$