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09.29.2004 at 05:57PM PDT, ID: 21150355
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Intersection point for two lines, given endpoints.

Tags: lines, two, intersection, point
I need code to find the intersection point of two lines given the endpoints of both lines.  Tried Liang-Barsky algorithm, with no luck.  I'm using C/C++ and openGL.

I did a search of this website first to see if my question had been previously answered, but the resulting webpages, just read: 404: page not found - so if this has been answered in the past, it's not on the site now.

Thanks for your help,
Ethan Deyo
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Question Stats
Zone: Programming
Question Asked By: etrek
Solution Provided By: str_ek
Participating Experts: 2
Solution Grade: A
Views: 104
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09.30.2004 at 08:06AM PDT, ID: 12190737

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10.01.2004 at 11:05AM PDT, ID: 12202518

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10.25.2004 at 09:04AM PDT, ID: 12401580

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10.29.2004 at 02:21AM PDT, ID: 12442507

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09.30.2004 at 08:06AM PDT, ID: 12190737
if lines in this contexts are line sections for 2d space the exact solution comes from deriving the common/intersect point of two line equations

so for 2d ( extra dimmentsion just add an extra param for extra coordinate)

if y=a*x+b and you got the coords of two point belonging to that line like (x1,y1) (x2,y2)

you can derive the a and b params :

y1=a*x1+b
y2=a*x2+b

b=y1-a*x1
y2=a*x2+y1-a*x1 => a=(y2-y1)/(x2-x1) => b=y1-(y2-y1)/(x2-x1)*x1

when you have deriven parameters (for both line - just like the example above )
you can tell that you seek an intersection of
y=a1*x+b1 line, and
y=a2*x+b2 line

x and y of this point is the common point of both lines, so it's the point that will compute both equatuins

yi=a1*xi+b1
yi=a2*xi+b2

a1*xi+b1=a2*xi+b2 => xi=(b2-b1)/(a1-a2)
yi=a1*(b2-b1)/(a1-a2)+b1

oh.. but you know the exact a1,a2,b1,b2 and you can make the equations more useful
just paste the a1 b1 etc values into solution

so.. if 1. line is (x11,y11;x12,y12) second (x21,y21;x22,y22)

xi=(b2-b1)/(a1-a2)
yi=a1*(b2-b1)/(a1-a2)+b1

xi=(y21-(y22-y21)/(x22-x21)*x21-y11-(y12-y11)/(x12-x11)*x11)/((y12-y11)/(x12-x11)-(y22-y21)/(x22-x21))
yi=(y12-y11)/(x12-x11)*(b2-b1)/((y12-y11)/(x12-x11)-(y22-y21)/(x22-x21))+y11-(y12-y11)/(x12-x11)*x11

it will return values or infinties (division by zero - so make code check for 0 in divisions!)
values are the common, and infiniteis (or 0 - in division as you would detect it) are for paralel lines !
ok, but that's not all!

you need to check if your intersection point belongs to your line sections, cos this is common somution of lines - not lines sections intersect !

this is done petty eazyly!

given points are guaranted to belong to the line your line secion lays on, so all you do is check the bound ...

if xi is between x11 and x12 , and yi is between y11 and y12 that means this point belongs to the line sections decribed by these points! so, alll you do is check the boundaries...

if your interscection pooint lays within boundaries of both line sections it is line sections intersection!

if you point lays beond boundaries that means lines intersect (arent pararel) , but sections dont

taht's all but' it


Accepted Solution
 
10.01.2004 at 11:05AM PDT, ID: 12202518
 
10.25.2004 at 09:04AM PDT, ID: 12401580
No comment has been added to this question in more than 21 days, so it is now classified as abandoned.

I will leave the following recommendation for this question in the Cleanup topic area:
   Split: str_ek {http:#12190737} & ozo {http:#12202518}

Any objections should be posted here in the next 4 days. After that time, the question will be closed.

Venabili
EE Cleanup Volunteer
 
10.29.2004 at 02:21AM PDT, ID: 12442507
Forced accept

modulo
Community Support Moderator
 
 
20080236-EE-VQP-29