I was wondering... How do you find the point where two lines intersect? This is assuming you have the coordinates (x1,y1,x2,y2) for the lines? If someone would help me with this I would be very grateful, as I am working on a Physics engine for my game that involves a bouncing ball (Hasbro PLEASE don''t sue me!!!!!). Thanks. Brent Robinson "What if this is as good as it gets?"

Any maths textbook will explain this. Anyway from the top of my head, a line is made up of 2 points. So (x1,y1) & (x2,y2) will only make up one line.

Anyway, lets assume you have two lines:
y1 = m1x1 + b1 and
y2 = m2x2 + b2

Now, assuming the 2 lines cross:
solve the y''s and the x''s by using the simultaneous equation.

Right?

Thanks! I think I have it now... Wait. How would you find the y-intercept for the lines? I know this is probably basic linear algebra, but I decided to go straight to Geometry my freshman year of High School. That said I am still a freshman if anyone is still confused.

Brent Robinson
"What if this is as good as it gets?"

to find the y-intercept you will have to make x=0;

so if you have this line equation:

y = 3x+7 then the y-intercept would be y= 7; (since x=0)

Though it technically falls under the domain of linear algebra, this is often referred to simply as 'Algebra'.

y= m*x + b



m1 = y1 / x1; //rise over run
b1 = y1 - m1 * x1; //substitute a point, any point will do

m2 = y2 / x2;
b2 = y2 - m2 * x2;

so we have:

y = m1*x + b1
y = m2*x + b2

so long as (m1<>m2 xor b1<>b2) a solution exist (if m1==m2 && b1==b2 the are the same line)

m2*x + b2 = m1*x + b1
m2*x - m1*x = b1 - b2
(m2 - m1)*x = b1 - b2
x = (b1 - b2) / (m2 - m1)

substitute x to yield
y = m1 * ((b1 - b2)/(m2 - m1)) + b1

the point defined by the equations for x & y above is the point of intersection.

...
Did I just do your homework for you?


Edited by - Magmai Kai Holmlor on November 8, 2000 12:28:29 AM

Ok, above have explained it pretty well, but I''ll give it try as well

We all know the basic equaion of a line i 2D:

y = k*x + m;

where[k] = y/x deltay(xsteps per step in y-direction)
and [m] = the value of y when x=0

ok, say we have two lines.

1: y = 2x + 2;
2: y = -x + 5;

Ok, substitute away Y : -x + 5 = 2x + 2;
to find the x coordinate then there should be
a soulution for this equation, lets try...

3 = 3x;
x = 1;

Now, when you know on which x they intersect, it''s easy
to find the y. Just plug it in to any of the two equations.
y = 2*1 +2;
y = 4;

yep, they intersected at 1,4.

Lets take two paralell lines and do the same task.

y = 2x + 1;
y = 2x + 5;

These two should not intersect (hopefully )

lets give it a try:
2x + 1 = 2x + 5;
0 = 4

Nop, just as we wanted, now it is just up to you implementing this in C/C++ or whatever you want, its 00:41, so I cant really think clearly anymore

/ Tooon

Hey, Thank you sooo much guys! That really helped in my game''s physics clipping, where I am trying to find the point where the ball''s path intersects a leg of a rectangle. I then just check to make sure the point is contained within those segments. In case your wondering, I am pretty new to programming and just learning the basics by making a similar to BREAKOUT. Thanks again!

Brent Robinson
"What if this is as good as it gets?"

Sorry guys, but one more thing. How do I compute the intersection of 2 lines where one is either Horizontal or vertical. I''m sorry, but I can''t seem to figure anything out today. Thanks.

Brent Robinson
"What if this is as good as it gets?"

horizontal is easy, the same as above

verticle is a problem

special case i guess...
if one line is verticle, and the other isn''t they intersect, and the x value is the value of the verticle line, feed into the other line''s equation to get y value.

if both are verticle, and b1=b2, then they are the same line...


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