Re: point on line
- Posted by tone.skoda at siol.net Feb 25, 2002
- 380 views
-- check if p is on the line defined by p1 and p2 function point_on_line(sequence p, sequence p1, sequence p2) if equal(p,p1) or equal(p,p2) then return 1 end if -- ADD THIS: -- if x's are same then should -- also y's be same, for point to be on line if p1[1]-p[1] = 0 then return 0 end if if p1[1]-p2[1] = 0 then -- line is vertical if p[0] = p1[0] then -- x's are same and line is vertical return 1 else -- x's are not same and line is vertical return 0 end if end if -- END ADD THIS. return ( (p1[2]-p[2])/(p1[1]-p[1]) ) = ( (p1[2]-p2[2])/(p1[1]-p2[1])) end function ----- Original Message ----- From: <bensler at mail.com> To: "EUforum" <EUforum at topica.com> Sent: Tuesday, February 26, 2002 2:20 AM Subject: point on line > > This function will fail with 'divide by zero' errors > > Can I fix this easily? I'm stumped. > > -- check if p is on the line defined by p1 and p2 > function point_on_line(sequence p, sequence p1, sequence p2) > if equal(p,p1) or equal(p,p2) then return 1 end if > return ( (p1[2]-p[2])/(p1[1]-p[1]) ) = ( (p1[2]-p2[2])/(p1[1]-p2[1])) > end function > > Chris > > > >