Transform a triangle to another triangle

Question

Hi i am trying to create the affine transform that will allow me to transform a triangle into another one. What i have are the coordinates for the 2 triangles. Can you help me? Following the answer by Adam Rosenfield i came up with this code in case anyone is bored to solve the equation himself : ``` public static AffineTransform createTransform(ThreePointSystem source, ThreePointSystem dest) { double x11 = source.point1.getX(); double x12 = source.point1.getY(); double x21 = source.point2.getX(); double x22 = source.point2.getY(); double x31 = source.point3.getX(); double x32 = source.point3.getY(); double y11 = dest.point1.getX(); double y12 = dest.point1.getY(); double y21 = dest.point2.getX(); double y22 = dest.point2.getY(); double y31 = dest.point3.getX(); double y32 = dest.point3.getY(); double a1 = ((y11-y21)*(x12-x32)-(y11-y31)*(x12-x22))/ ((x11-x21)*(x12-x32)-(x11-x31)*(x12-x22)); double a2 = ((y11-y21)*(x11-x31)-(y11-y31)*(x11-x21))/ ((x12-x22)*(x11-x31)-(x12-x32)*(x11-x21)…

Accepted Answer

I'm going to assume you're talking about 2D here. An affine transformation matrix has 9 values in it:

```
    | a1 a2 a3 |
A = | a4 a5 a6 |
    | a7 a8 a9 |
```

There are 3 input vertices `x1`, `x2`, and `x3`, which when transformed should become `y1`, `y2`, `y3`. However, since we're working in homogeneous coordinates, applying `A` to `x1` doesn't necessarily give `y1` -- it gives a multiple of `y1`. So, we also have the unknown multipliers `k1`, `k2`, and `k3`, with the equations:

```
A*x1 = k1*y1
A*x2 = k2*y2
A*x3 = k3*y3
```

Each of those is a vector, so we really have 9 equations in 12 unknowns, so the solution is going to be underconstrained. If we require that `a7=0`, `a8=0`, and `a9=1`, then the solution will be unique (this choice is natural, since it means if the input point is (`x`, `y`, 1), then the output point will always have homogeneous coordinate 1, so the resulting transform is just a 2x2 transform plus a translation).

Hence, this reduces the equations to:

```
a1*x11 + a2*x12 + a3 = k1*y11
a4*x11 + a5*x12 + a6 = k1*y12
                   1 = k1
a1*x21 + a2*x22 + a3 = k2*y21
a4*x21 + a5*x22 + a6 = k2*y22
                   1 = k2
a1*x31 + a2*x32 + a3 = k3*y31
a4*x31 + a5*x32 + a6 = k3*y32
                   1 = k3
```

So, `k1` = `k2` = `k3` = 1. Plugging these in and converting to matrix form yields:

```
| x11 x12   1   0   0   0 |   | a1 |   | y11 |
| x21 x22   1   0   0   0 |   | a2 |   | y21 |
| x31 x32   1   0   0   0 | * | a3 | = | y31 |
|   0   0   0 x11 x12   1 |   | a4 |   | y12 |
|   0   0   0 x21 x22   1 |   | a5 |   | y22 |
|   0   0   0 x31 x32   1 |   | a6 |   | y32 |
```

Solving this 6x6 system of equations yields you your affine transformation matrix `A`. It will have a unique solution if and only if the 3 points of your source triangle are not collinear.

Transform a triangle to another triangle

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