what is the mechanism of intersection between two lines ?

I got this code from internet, It works for intersection of two lines. but i dont understand how it works ? Please anyone explain it.

x1  = [v(1,1) v(1,2)]; %vertical line X intersection point.
y1  = [v(2,1) v(2,2)];%vertical line Y intersection point.
%line2
x2 = [h(1,1) h(1,2)];%Horizontal line X intersection point.
y2 = [h(2,1) h(2,2)];%Horizontal line Y intersection point.
%fit linear polynomial
p1 = polyfit(x1,y1,1);
p2 = polyfit(x2,y2,1);
%calculate intersection
x_intersect = fzero(@(x) polyval(p1-p2,x),3);
y_intersect = polyval(p1,x_intersect);
P(1)=x_intersect;
P(2)=y_intersect;

Answer 1

[David Goodmanson]

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.

[John D'Errico]

+1. Exactly. While you can use tools like polyfit, fzero, etc., to solve the problem. WHY? Don't use a Mack truck to take a pea to Boston, when just very little basic mathematics is sufficient to solve the problem.

The code that you got off the internet is worth just what you paid for it: nothing. It is crappy, bad, lazy code. Code that was written without real thought, solving a problem using overkill, for only an approximate solution since it uses fzero!

[sufian ahmed]

i didnot understand how it works. can u provide me a code with explanation.

[Jan]

What is the relation between this picture and the question?

[sufian ahmed]

@jan i want to find this intersections, showed in image. But how can i do the code in matlab. since i have:

Line 1: [x1 y1] [x2 y2] Line 2: [x3 y3] [x4 y4]

[Image Analyst]

Well, do you have an image, OR do you have equations of lines? Exactly what are you starting with?

Answer 2

[Image Analyst]

See

https://blogs.mathworks.com/steve/2016/04/12/intersecting-curves-that-dont-intersect/

http://blogs.mathworks.com/loren/2011/08/29/intersecting-lines/?s_tid=srchtitle

http://blogs.mathworks.com/pick/2011/09/09/detect-curve-intersections-quickly-and-easily/?s_tid=srchtitle