# How to compute a 2x2 Homography out of 3 corresponding points?

In 1D projective geometry, I want to compute the 2x2 Homography matrix $H$ (in homogeneous coordinates), given 3 pairs of corresponding points.

i.e. I want to find H such that: $$\left(\begin{array}{cc} h_{11} & h_{12}\\ h_{21} & h_{22}\end{array}\right)\left(\begin{array}{ccc} 0 & a & a+b\\ 1 & 1 & 1 \end{array}\right) = \left(\begin{array}{ccc} 0 &a' &a'+b'\\ 1 & 1 & 1 \end{array}\right).$$

However, I've got 6 equations here and only 3 unknowns. (dof(H) = 4 elements less one for scaling = 3).

I thought about 3 scaling factors that would add up to 6 unknowns, s.t. we would have a unique solution. But how exactly do I insert the scaling factors into the matrices and how can I compute H then?

Do you have a clue?

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## 2 Answers

I believe the result should be in the form $$\left(\begin{array}{ccc} 0 & k_2a' & k_3(a'+b')\\ k_1 & k_2 & k_3 \end{array}\right)$$ with $k_1,k_2,k_3$ nonzero, to account for the homogeneous coordinates. Remember that the projective point $[x:y]$ is the same as the point $[kx:ky]$. So the middle answer doesn't have to look like $[a':1]$, it can be any of the other representations of the point, hence $[k_2a':k_2]$. That gives you three "scaling" factors. Is that what you were hoping for?

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Your answer is mathematically correct, however I figured out another way to solve this equation, which leads to a simpler result.

I applied the technique for 2D projective geometry, which is described here to the 1D case and it works out fine.

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