It looks like you're new here. If you want to get involved, click one of these buttons!

- 140.9K All Categories
- 104.5K Programming Languages
- 6.4K Assembler Developer
- 1.8K Basic
- 39.7K C and C++
- 4.2K C#
- 7.9K Delphi and Kylix
- 4 Haskell
- 9.6K Java
- 4.1K Pascal
- 1.3K Perl
- 1.9K PHP
- 507 Python
- 48 Ruby
- 4.3K VB.NET
- 1.6K VBA
- 20.8K Visual Basic
- 2.6K Game programming
- 310 Console programming
- 88 DirectX Game dev
- 1 Minecraft
- 109 Newbie Game Programmers
- 2 Oculus Rift
- 8.9K Applications
- 1.8K Computer Graphics
- 726 Computer Hardware
- 3.4K Database & SQL
- 521 Electronics development
- 1.6K Matlab
- 627 Sound & Music
- 256 XML Development
- 3.3K Classifieds
- 191 Co-operative Projects
- 180 For sale
- 189 FreeLance Software City
- 1.9K Jobs Available
- 600 Jobs Wanted
- 201 Wanted
- 2.9K Microsoft .NET
- 1.7K ASP.NET
- 1.1K .NET General
- 3K Miscellaneous
- 3 Join the Team
- 2 User Profiles
- 353 Comments on this site
- 59 Computer Emulators
- 1.8K General programming
- 178 New programming languages
- 608 Off topic board
- 168 Mobile & Wireless
- 42 Android
- 124 Palm Pilot
- 335 Multimedia
- 151 Demo programming
- 184 MP3 programming
- 0 Bash scripts
- 18 Cloud Computing
- 52 FreeBSD
- 1.7K LINUX programming
- 367 MS-DOS
- 0 Shell scripting
- 320 Windows CE & Pocket PC
- 4.1K Windows programming
- 889 Software Development
- 405 Algorithms
- 68 Object Orientation
- 86 Project Management
- 88 Quality & Testing
- 234 Security
- 7.5K WEB-Development
- 1.8K Active Server Pages
- 61 AJAX
- 2 Bootstrap Themes
- 55 CGI Development
- 19 ColdFusion
- 222 Flash development
- 1.4K HTML & WEB-Design
- 1.4K Internet Development
- 2.2K JavaScript
- 33 JQuery
- 285 WEB Servers
- 119 WEB-Services / SOAP

Welcome to the new platform of Programmer's Heaven! We apologize for the inconvenience caused, if you visited us from a broken link of the previous version. The main reason to move to a new platform is to provide more effective and collaborative experience to you all. Please feel free to experience the new platform and use its exciting features. Contact us for any issue that you need to get clarified. We are more than happy to help you.

The code would look something like the following:

[CODE]RECT rect_array[n];

...

bool DoesLineIntersectAny(RECT line)

{

for(int i = 0; i < n; i++)

if(LineIntersects(rect_array[i], line)

return true;

return false;

}

[/CODE]Unfortunately I cannot find a function such as "LineIntersects" in any standard libraries or API's. Any help would be greatly appreciated.

About & Contact / Terms of use / Privacy statement / Publisher: Lars Hagelin

Programmers Heaven articles / Programmers Heaven files / Programmers Heaven uploaded content / Programmers Heaven C Sharp ebook / Operated by CommunityHeaven LLC

© 1997-2013 Programmersheaven.com - All rights reserved.

## Comments

75MemberFirst, determine an equation for each line. You can do this easily by working out the gradient of the line (increase in y / increase in x) and the find the constant of the equation, which will be the value of y when x = 0.

So say you have a line made up of 2 points with co-ordinates (10, 10) and (15, 20). The increase in y is 10 and the increase in x is 5, so the gradient is 10 / 5 = 2. When x = 0, y will be 5, so the constant in the equation is 5. This will give equation of the line as

y = 2x + 5

Now do the same for the second line. Assume the gradient is 4 and the constant is 13. This makes the second equation

y = 4x + 13

To find the point of intersection, you just need to find where the 2 equations are equal. Do this by combining the equations like this:

2x + 5

4x + 13

2x + 5 = 4x + 13

Rearrange this to make x on it's own on the left hand side:

2x + 5 = 4x + 13

-2x + 5 = 0x + 13

-2x + 0 = 0x + 8

-2x = 8

x = -4

This result means the 2 lines intersect when x is at -4

You can then substitute this value of x into any of the 2 equations to retrieve the y co-ordinate.

Actually the thing with straight lines is, there will ALWAYS be a point of intersection! The only exception of course is if the 2 lines are parallel, or they have the same gradient. To do a simple check of whether 2 lines will ever cross, check to see if their gradients are equal. I'm just assuming you wanted to know where they intersect.

Hope that solves your problem.

- Spam

0 · Vote Down Vote Up · Share on Facebook78Member- Spam

0 · Vote Down Vote Up · Share on Facebook621Member:

: First, determine an equation for each line. You can do this easily by working out the gradient of the line (increase in y / increase in x) and the find the constant of the equation, which will be the value of y when x = 0.

:

: So say you have a line made up of 2 points with co-ordinates (10, 10) and (15, 20). The increase in y is 10 and the increase in x is 5, so the gradient is 10 / 5 = 2. When x = 0, y will be 5, so the constant in the equation is 5. This will give equation of the line as

: y = 2x + 5

:

: Now do the same for the second line. Assume the gradient is 4 and the constant is 13. This makes the second equation

: y = 4x + 13

:

: To find the point of intersection, you just need to find where the 2 equations are equal. Do this by combining the equations like this:

:

: 2x + 5

: 4x + 13

:

: 2x + 5 = 4x + 13

:

: Rearrange this to make x on it's own on the left hand side:

:

: 2x + 5 = 4x + 13

: -2x + 5 = 0x + 13

: -2x + 0 = 0x + 8

: -2x = 8

: x = -4

:

: This result means the 2 lines intersect when x is at -4

: You can then substitute this value of x into any of the 2 equations to retrieve the y co-ordinate.

:

: Actually the thing with straight lines is, there will ALWAYS be a point of intersection! The only exception of course is if the 2 lines are parallel, or they have the same gradient. To do a simple check of whether 2 lines will ever cross, check to see if their gradients are equal. I'm just assuming you wanted to know where they intersect.

:

: Hope that solves your problem.

The lines mentioned do not always intersect. They have a start and end point and therefore have a length.

If you find the X for which two lines intersect, you'll have to check if that value does not fall outside of both lines' starting and ending X-values.

Greets,

Eric Goldstein

http://www.gvh-maatwerk.nl

- Spam

0 · Vote Down Vote Up · Share on Facebook1,018Member: :

: : First, determine an equation for each line. You can do this easily by working out the gradient of the line (increase in y / increase in x) and the find the constant of the equation, which will be the value of y when x = 0.

: :

: : So say you have a line made up of 2 points with co-ordinates (10, 10) and (15, 20). The increase in y is 10 and the increase in x is 5, so the gradient is 10 / 5 = 2. When x = 0, y will be 5, so the constant in the equation is 5. This will give equation of the line as

: : y = 2x + 5

: :

: : Now do the same for the second line. Assume the gradient is 4 and the constant is 13. This makes the second equation

: : y = 4x + 13

: :

: : To find the point of intersection, you just need to find where the 2 equations are equal. Do this by combining the equations like this:

: :

: : 2x + 5

: : 4x + 13

: :

: : 2x + 5 = 4x + 13

: :

: : Rearrange this to make x on it's own on the left hand side:

: :

: : 2x + 5 = 4x + 13

: : -2x + 5 = 0x + 13

: : -2x + 0 = 0x + 8

: : -2x = 8

: : x = -4

: :

: : This result means the 2 lines intersect when x is at -4

: : You can then substitute this value of x into any of the 2 equations to retrieve the y co-ordinate.

: :

: : Actually the thing with straight lines is, there will ALWAYS be a point of intersection! The only exception of course is if the 2 lines are parallel, or they have the same gradient. To do a simple check of whether 2 lines will ever cross, check to see if their gradients are equal. I'm just assuming you wanted to know where they intersect.

: :

: : Hope that solves your problem.

:

: The lines mentioned do not always intersect. They have a start and end point and therefore have a length.

: If you find the X for which two lines intersect, you'll have to check if that value does not fall outside of both lines' starting and ending X-values.

:

:

: Greets,

: Eric Goldstein

: http://www.gvh-maatwerk.nl

:

:

:

[green]

The domain of line y=mx+b is the set of reals. It's range is the set of all reals, so it could go on to -infinity to +infinity. What you are describing is the distance between two points which is only part of a line or also called a line segment.

[/green]

- Spam

0 · Vote Down Vote Up · Share on Facebook