Hi sufian,
Well, you can find all kinds of code on the internet, and on this forum most people are more likely to provide code ideas than try to unravel code by some third party. Especially in a case like this which uses two polyfits, an fzero and a polyval to solve this problem, all of which are not necessary.
Assume there is a line defined by endpoints a and b and another defined by endpoints c and d. Each point is given by a 2x1 column vector such as [ax; ay]. Any point along the line ab (or its extension off the ends) is described by p = a + alpha*(b-a) for some scalar alpha, similarly for line cd. The intersection point occurs when
a + alpha*(b-a) = c + beta*(d-c)
for some unique alpha and beta which you have to solve for. If you put this into matrix form you can arrive at
[(b-a) (d-c)]*lambda = (c-a)
where [...] is 2x2, and lambda is the vector [alpha; -beta] as you can check. (In this equation the top row is x components and the bottom row is y components). The solution is simply
lambda = [(b-a) (d-c)]\(c-a);
then p = a + lambda(1)*(b-a)
and also p = c - lambda(2)*(d-c)
as a check.
