I have two line segments, one with points $P_1=(x_1,y_1,1), P_2=(x_2,y_2,1)$ and other with points $P_3=(x_3,y_3,1), P_4=(x_4,y_4,1)$. I need to find transform matrix
$$M=\left(\begin{array}{ccc}
a & b & c\\
d & e & f\\
0 & 0 & 1
\end{array}\right)$$
so that
$$M\cdot P_1 = P_3$$
$$M\cdot P_2 = P_4.$$
Also, transform should equally scale points along X and Y axis.
I have tried it in following way:
$$M=
\left(\begin{array}{ccc}
\cos\alpha & -\sin\alpha & 0\\
\sin\alpha & \cos\alpha & 0\\
0 & 0 & 1
\end{array}\right)
\left(\begin{array}{ccc}
1 & 0 & h\\
0 & 1 & v\\
0 & 0 & 1
\end{array}\right)
\left(\begin{array}{ccc}
s & 0 & 0\\
0 & s & 0\\
0 & 0 & 1
\end{array}\right).$$
Then I have system of four equations in four unknowns, which is not trivial to solve. I have tried to solve it with Mathematica, however, it gives answer in very messy form.
I'm wondering, is there any more elegant way to solve this problem and short expressions for matrix coefficients ($a,b,c,d,e,f$)?
Best Answer
General Idea
You were on the right track with those matrices! Instead of trying to solve the system of equations in your $M$ matrix, it's easier to figure out each transformation individually and then just multiply the results together.
Let's start with a (slightly) simplified version, where the initial line segment lies along the $x$-axis, with one point at the origin. With that assumption, it becomes easier to visualize how we could change one segment into the other:
(Your $M$ matrix basically does this- although it scales, translates, then rotates, which won't end up with the right result since you're rotating about the origin.)
Scale
The scale factor is just the ratio of the two lengths ($\frac{l_2}{l_1}$), since you want the final length to be $l_2$ and $l_1\times\frac{l_2}{l_1}=l_2$. To scale points equally along both axes, you use the scaling matrix you had with the scaling factor:
$$\left(\begin{array}{ccc} \frac{l_2}{l_1} & 0 & 0\\ 0 & \frac{l_2}{l_1} & 0\\ 0 & 0 & 1 \end{array}\right)$$
To calculate $l_1$ and $l_2$, we can use the distance formula ($\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$).
Rotate
Because of our simplifying assumption that the initial segment starts at $(0,0)$ and lies along the $x$-axis, figuring out the rotation matrix is just a matter of figuring out the angle the second segment makes with the $x$-axis. To do this, we can use atan2, which is just like $tan^{-1}$ but takes the signs/quadrants into account.
$atan2$ works with vectors, so we need to shift the second segment to the origin before doing the calculation. This is done by subtracting the coordinates of one point of the segment from the other.
For example, for a segment with points $a=(a_x,a_y)$ and $b=(b_x,b_y)$, the angle it makes with the $x$-axis is $\theta=\text{atan2}(b_y-a_y, b_x-a_x)$, which can be plugged into your transformation matrix:
$$\left(\begin{array}{ccc} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{array}\right)$$
Translate
At this point, the translation is easy, we just need to shift $(0,0)$ to the corresponding point on the final segment:
$$\left(\begin{array}{ccc} 1 & 0 & x_3\\ 0 & 1 & y_3\\ 0 & 0 & 1 \end{array}\right)$$
Putting Them Together
The final transformation matrix is just the three we've derived combined:
$$ M= \left(\begin{array}{ccc} 1 & 0 & x_3\\ 0 & 1 & y_3\\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} \frac{l_2}{l_1} & 0 & 0\\ 0 & \frac{l_2}{l_1} & 0\\ 0 & 0 & 1 \end{array}\right) $$
But... What If The First Segment Isn't In A Nice Place?
We can put it there! Using the same techniques we've already used, all you have to do is: