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Techno-Logic
Jul 14th, 2002, 03:17 PM
I have been trying to solve this problem for a long time, but unfortunately couldn`t find any solution.

In the picture, you see, there are the two lines A-B and D-E. Suppose, their coordinates are defined and known. They intersect at the point C, and I must get the coordinates of this point.

I have tried to create some equalities, but couldn`t have a success.

Would you please share your ideas with me, to solve this.

Thanks in advance.

snakeeyes1000
Jul 14th, 2002, 03:37 PM
put them both in slope intercept form. Follow the example below.

equation #1
--------------
SLOPE=1
Y-INTERCEPT=-3
-------------
y=x-3

equation #2
---------------
SLOPE=-1
Y-INTERCEPT=0
--------------
y=-x+0

now look at the two equations like this.

y=+x-3
y=-x+0

We then add the equations like so
2y=-3

hence we get
y=-3/2

subtract and we get
0=2x-3 -> 2x=3 -> x=3/2

so they intersect at (3/2,-3/2)

Guv
Jul 14th, 2002, 04:16 PM
What information do you have about the two lines?

No matter how the lines are determined, you should be able to set up two equations in two unknowns. Susch systems are easy to solve.

Techno-Logic
Jul 14th, 2002, 04:51 PM
Firstly, I would like to thank you for your interest and replies

In this problem, I only know the start and the end points of both of the lines.

I uploaded the picture of these lines with defined coordinates, please take a look.

Excuise me, please, but I`m a bit confused, and can`t really understand these equations. Would you please give a little more details.

Thanks again.

bugzpodder
Jul 14th, 2002, 09:41 PM
first calculate the slope of both lines by taking the two points that forms a line -- call them (x1,y1),(x2,y2) and do
slope(m)=(y2-y1)/(x2-x1)

then you put them into y=mx+b form (personally i like the standard form better but slope-y intercept form is easier for starters). in y=mx+b, m is the slope. (you are trying to find the equation of the line that joins both points) so plug in m, and plug in (x1,y1) or (x2,y2) and solve for b. do the same for the other equation. now you have something like:
y=ax+b
y=cx+d
where a,b,c,d are constant values
equate them together to get:
ax+b=cx+d
and solve for x
plug it back into either equation to get the y value
the solution (x,y) will be the point of intersection.

Techno-Logic
Jul 15th, 2002, 05:24 AM
Thanks a billion for your help and sharing your knowledge with me, Bugzpodder.
Also thanks a lot for Snakeeyes1000 and Guv for your interest, help and replies.

Techno-Logic
Jul 15th, 2002, 05:37 AM
Thanks a billion for your help and sharing your knowledge with me, Bugzpodder.
Also thanks a lot for Snakeeyes1000 and Guv for your interest, help and replies.

Gandalf_Grey_
Jul 16th, 2002, 06:29 PM
another way to solve the first problem (i hate snake eyes way) is you set the y's equal and also set the rest of the equations equal because y =y and x-3 = -x, then solve for x. then plug the x number back into either of the original equations and solve for them

transcendental
Jul 16th, 2002, 08:12 PM
General Solution to Intersection of 2 Lines
============================

Calculate the gradient of the 2 lines(m)

y=mx+C

Get any points (x,y) and substitute to get C for each line.

Rearrange the 2 equations.

y-mx=C----eq(1)for first line
y-mx=C----eq(2)for second line

Solve for the simultaneous equations, using cramer's rule.

U will get the intersection points, as long as the lines are not parallel.

How to know they are parallel?, the determinant for

|y1,-mx1|
|y2,-mx2|

is 0, meaning they are parallel,no need to test further, else u get 'divide by zero error'.

Gordon Bennett
Jul 17th, 2002, 07:37 AM
You speak of your lines having start and end points. In two dimensions (like on a graph), two lines which have ends to them don't have to intersect. A line from the origin to the point (1,1) doesn't intersect with a line from (0,3) to (0,4). Do yourself a drawing and see for yourself (with your lines, not mine).