Area between two lines inside the square [-1,+1] x [-1,+1]

I'm working on a project in Matlab and need to find the area between two lines inside of the square [-1,+1]x[-1,+1] intersecting in a point (xIntersection,yIntersection). So the idea is to subtract the two lines and integrate between [-1, xIntersection] and [xIntersection, +1], sum the results and if it's negative, change its sign.

For details on how I find the intersection of the two lines check this link.

I'm using Matlab's function integral(), here a snippet of my code:

xIntersection = ((x_1 * y_2 - y_1 * x_2) * (x_3 - x_4) - (x_1 - x_2) * (x_3 * y_4 - y_3 * x_4) ) / ((x_1 - x_2) * (y_3 - y_4) - (y_1 - y_2) * (x_3 - x_4));

d = @(x) g(x) - f(x);
result = integral(d, -1, xIntersection) - int( d, xIntersection, 1)
if(result < 0),
    result = result * -1;
end

Note that I defined previously in the code g(x) and f(x) but haven't reported it in the snippet.

The problem is that I soon realized that the lines could intersect either inside or outside of the square, furthermore they could intersect the square on any of its sides and the number of possible combinations grows up very quickly.

I.e.:

These are just 4 cases, but considering that f(+1), f(-1), g(+1), g(-1) could be inside the interval [-1,+1], above it or under it and that the intersection could be inside or outside of the square the total number is 3*3*3*3*2 = 162.

Obviously in each case the explicit function to integrate in order to get the area between the two lines is different, but I can't possibly think of writing a switch case for each one.

Any ideas?

I think my answer to your previous question still applies, for the most part.

If you want to compute the area of the region bounded by the smaller angle of the two lines and the boundaries of the square, then you can forget about the intersection, and forget about all the different cases.

You can just use the fact that

• the area of the square S is equal to 4

• the value of this integral

  A = quadgk(@(x) ...
      abs( max(min(line1(x),+1),-1) - max(min(line2(x),+1),-1) ), -1, +1);
  

  gives you the area between the lines (sometimes the large angle, sometimes the small angle)

• the value of of min(A, S-A) is the correct answer (always the small angle).