Get the points of intersection from 2 rectangles

Question

Let say that we have two rectangles, defined with their bottom-left and top-right corners. For example: rect1 (x1, y1)(x2, y2) and rect2 (x3, y3)(x

Answer 1

Let's say that a box has a radius X and radius Y (I know it has not but this term is useful here).

You will have:

rect1_x_radius = (x2-x1)/2
rect1_y_radius = (y2-y1)/2

and

rect2_x_radius = (x4-x3)/2
rect2_y_radius = (y4-y3)/2

Now if rect middle points are further away than sum of their radiuses in appropriate direction - they do not collide. Otherwise they do - this hint should suffice.

You should be now able to finish your assignment.

UPDATE:

OK - let's solve it for 1D - later you'll solve it for 2D. Look at this piece of ... art ;-)

You see 2 segments - now some calculations:

rA = (maxA-minA) / 2
rB = (maxB-minB) / 2

midA = minA + rA
midB = minB + rB

mid_dist = |midA - midB|

Now how to check if collision occurs? As I said if sum of 'radiuses' is less than segments' distance - there is no collision:

if ( mid_dist > fabs(rA+rB) )
{
    // no intersection
}
else
{
    // segments intersect
}

Now it is your work to calculate intersection / common part in 1D and 2D. It is up to you now (o ryou can read Andrey's answer).

Here is the same situation but in 2D - two 1D situations: