One solution is to use the equations derived in this tutorial for finding the intersection point of two lines in 2-D (update: this is an internet archive link since the site no longer exists). You can first create two matrices: one to hold the x coordinates of the line endpoints and one to hold the y coordinates.
x = [0 0; 6 6]; %# Starting points in first row, ending points in second row
y = [0 6; 6 0];
The equations from the above source can then be coded up as follows:
dx = diff(x); %# Take the differences down each column
dy = diff(y);
den = dx(1)*dy(2)-dy(1)*dx(2); %# Precompute the denominator
ua = (dx(2)*(y(1)-y(3))-dy(2)*(x(1)-x(3)))/den;
ub = (dx(1)*(y(1)-y(3))-dy(1)*(x(1)-x(3)))/den;
And you can now compute the intersection point of the two lines:
xi = x(1)+ua*dx(1);
yi = y(1)+ua*dy(1);
For the example in the question, the above code gives xi = 3
and yi = 3
, as expected. If you want to check that the intersection point lies between the endpoints of the lines (i.e. they are finite line segments), you just have to check that the values ua
and ub
both lie between 0 and 1:
isInSegment = all(([ua ub] >= 0) & ([ua ub] <= 1));
A couple more points from the tutorial I linked to above:
- If the denominator
den
is 0 then the two lines are parallel.
- If the denominator and numerator for the equations for
ua
and ub
are 0 then the two lines are coincident.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…