intersecting lines
Simple question:
line AB intersects line CD at point P.
solve P.
[Edited by - Dauid on January 22, 2009 2:14:24 PM]
Well,
AB : ax + by = c ( basic math, this is the ecuation of a line )
CD : dx + ey = f
then you just make a system of the two ecuations and the resulting x's and y's are the coordonates of P
*later edit*
let's presume line AB is defined as : 3x + 2y = 1
CD is defined as : 2x + y = 0
{ 2y = 1 - 3x | *(-2)
{ y = - 2x | *3
{ -4y = -2 + 6x
{ 3y = -6x
----------------
-y = -2 => |y = 2|
and now for the x
4 = 1 - 3x
-3x = 3
|x = -1;|
so there you go, the point of intersection between these two lines is P ( -1, 2 )
*/later edit*
Hope I made some sense :P
AB : ax + by = c ( basic math, this is the ecuation of a line )
CD : dx + ey = f
then you just make a system of the two ecuations and the resulting x's and y's are the coordonates of P
*later edit*
let's presume line AB is defined as : 3x + 2y = 1
CD is defined as : 2x + y = 0
{ 2y = 1 - 3x | *(-2)
{ y = - 2x | *3
{ -4y = -2 + 6x
{ 3y = -6x
----------------
-y = -2 => |y = 2|
and now for the x
4 = 1 - 3x
-3x = 3
|x = -1;|
so there you go, the point of intersection between these two lines is P ( -1, 2 )
*/later edit*
Hope I made some sense :P
Quote:Original post by Dauid
Simple question:
line AB intersects line CD at point P.
define P.
You just did... P is the point where line AB intersects line CD. Perhaps you didn't mean "define", but something else?
hm, I should have mentioned I was trying to avoid the lineequation.
consider this example, you have a 4-sided polygon,
assign the diagonals to the polygon and solve where the diagonals intersect.
consider this example, you have a 4-sided polygon,
assign the diagonals to the polygon and solve where the diagonals intersect.
But you must have some sorts of coordinates for the points of the lines ?
In which case you can use the (y = mx+b) formula to find out the ecuation and move forward.
In which case you can use the (y = mx+b) formula to find out the ecuation and move forward.
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