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Jul 14th, 2002, 03:17 PM
#1
Thread Starter
Junior Member
Geometry Problem - Intersection
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.
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Jul 14th, 2002, 03:37 PM
#2
Hyperactive Member
put them both in slope intercept form. Follow the example below.
equation #1
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SLOPE=1
Y-INTERCEPT=-3
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y=x-3
equation #2
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SLOPE=-1
Y-INTERCEPT=0
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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)
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Jul 14th, 2002, 04:16 PM
#3
Frenzied Member
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.
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Jul 14th, 2002, 04:51 PM
#4
Thread Starter
Junior Member
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.
Last edited by Techno-Logic; Jul 14th, 2002 at 05:00 PM.
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Jul 14th, 2002, 09:41 PM
#5
Fanatic Member
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.
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Jul 15th, 2002, 05:24 AM
#6
Thread Starter
Junior Member
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.
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Jul 15th, 2002, 05:37 AM
#7
Thread Starter
Junior Member
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.
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Jul 16th, 2002, 06:29 PM
#8
Fanatic Member
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
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Jul 16th, 2002, 08:12 PM
#9
Hyperactive Member
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'.
Last edited by transcendental; Jul 16th, 2002 at 08:20 PM.
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Jul 17th, 2002, 07:37 AM
#10
Member
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).
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