geometry-challenged

[Jim Hurley]

>--
>
>Message: 2
>Date: Mon, 26 Jan 2004 06:57:50 -0500
>From: Jim Lyons <jimlyons at earthlink.net>
>Subject: Re: geometry-challenged
>To: use-revolution at lists.runrev.com
>Message-ID: <DDBA098E-4FF6-11D8-BEAE-000A95893982 at earthlink.net>
>Content-Type: text/plain; charset=US-ASCII; format=flowed
>
>Jim Hurley wrote in part:
>
>>  I'm not sure if this helps, but the function "intersection" below
>>  will return the intersection of any two lines.
>>  ...
>>
>>  function intersection p1,p2,pp1,pp2
>>      ...
>>
>>      return x & comma & y
>>  end intersection
>
>Someone already pointed out that this function as written still allows
>one divide by zero case.


Jim

Check. I think the following handler takes care of the infinity.

One practical way to deal with the parallel lines case is to make 
them converge to a point at a very great distance--see below. Or, of 
course, you could return a "Lines parallel" message.


function intersection p1,p2,pp1,pp2
   get the paramcount
   if it is 1 then
     put item 4 of p1 into pp2
     put item 3 of p1 into pp1
     put item 2 of p1 into p2
     put item 1 of p1 into p1
   end if
   put item 1 of p1 into x1
   put item 2 of p1 into y1
   put item 1 of p2 into x2
   put item 2 of p2 into y2
   put item 1 of pp1 into xp1
   put item 2 of pp1 into yp1
   put item 1 of pp2 into xp2
   put item 2 of pp2 into yp2
   --Check if the lines are parallel and deal with it somehow
   if (y2-y1)*(xp2-xp1) = (yp2-yp1)*(x2-x1) then add .0000001 to x2
   if x1 = x2 or xp1 = xp2 then
       if x2 = x1 then
       return x1&comma&yp2 + (x1-xp2)*(yp2-yp1)/(xp2-xp1)
     else
       return xp2&comma& y2 + (xp1-x2)*(y2-y1)/(x2-x1)
     end if
   end if
   put yp2 - y2 + x2*(y2-y1)/(x2 - x1) - xp2*(yp2 - yp1)/(xp2 - xp1) 
into numerator
   put ((y2 - y1)/(x2 - x1) -(yp2 - yp1)/(xp2-xp1)) into denom
   put numerator / denom into x
   put y2 + (x-x2) *(y2-y1)/(x2-x1) into y
   return x & comma & y
end intersection


Jim