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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.

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## 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