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.