solving two linear equation in program

Question

I have two lines UV and PQ, with U(15,10) V(50,25) and P(40,10) and Q(30,30).

and i am finding a point of intersection between these two points. I am forming two linear equations for that:

eq 1:

Ux + t(Vx-Ux) = Px +s(Qx-Px)

eq2

I want to solve these equations using C program to get the value of either t or s.

I used the t from second equation and substituted it in first equation to get a formula to find s. But it did not work out. How can I solve this in a program?

"I tried to drive to work, but it didn't work out. Do you have any suggestions?" What *exactly* did you try? What did you expect to happen? What happened? — David Schwartz, Sep 28 '13 at 04:49
[[a b][c d]][s t] = [A B] --> [[d -b] [-c a]][A B]/det() = [s t] — chux - Reinstate Monica, Sep 28 '13 at 04:49

score 1 · Answer 1 · answered Sep 28 '13 at 04:48

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Use this equation Intersection of 2 Lines. You can use either the long expanded form, or abstract out a function that calculates a 2x2 determinant. Be careful of using integers, you might overflow.

answered Sep 28 '13 at 04:48

DanielV

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this is for infinitely long lines . But i want to find intersection for two line segment. – Force.comBat Sep 28 '13 at 04:55
Then just check if the intersection-point's x and y are between the line segments' x and y. – DanielV Sep 28 '13 at 05:07

score 1 · Answer 2 · answered Sep 28 '13 at 04:57

1

int s,sa,sb;

Okay, sa is an integer.

sa=1/((Vx-Ux)*(Qy-Py)-(Qx-Px)*(Vy-Uy));

Since these are all integers, you're taking the reciprocal of an integer using integer math. The result is infinity if the integer is zero, one if it's one, negative one if it's negative one, and zero if it's anything else. So sa only has three possible values, or you crash.

Perhaps you should consider not using integer math? Maybe float sa; and sa = 1.0 / (....

answered Sep 28 '13 at 04:57

David Schwartz

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i get Sa as zero always. – Force.comBat Sep 28 '13 at 05:07
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@Rdvsm That makes perfect sense now, doesn't it. – David Schwartz Sep 28 '13 at 05:09
How can it be solved then? – Force.comBat Sep 28 '13 at 05:33
Look at my last paragraph. Don't use integer math. Use `float`s and floating point constants. – David Schwartz Sep 28 '13 at 05:33

score 0 · Answer 3 · answered Sep 28 '13 at 05:07

0

Find the gradient of UV and PQ

The gradient of UV is : m = (Vy-Uy)/(Vx-Ux). And then find c by using y = mx + c since we know the value of y, m, x. Do that step again for PQ.

After you have 2 equations, you can find the intersection point by using substitutions. And finally, apply that to your c code.

answered Sep 28 '13 at 05:07

david

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I dont want to use mx+c , Can u suggest a solution for above problem? – Force.comBat Sep 28 '13 at 05:33
Does the above formula is right ? – david Sep 28 '13 at 05:38

score 0 · Answer 4 · answered Sep 29 '13 at 00:56

OP's comment says "find intersection for two line segment".

AS OP provided pseudo-code, I assume C-like pseudo-code is wanted.

Change each line into a parametrized form P(t) = b + m*t (use double)
UV_m.x = V.x - U.x
UV_m.y = V.y - U.y
UV_b.x = U.x
UV_b.y = U.y
// same for PQ
// Now your have 2 2-dimensional equations.
// Puv(t1) = UV_b + UV_m * t1 and
// Ppq(t2) = PQ_b + PQ_m * t2
Assume Puv(t1) == Ppq(t2) for some t1,t2 (that is they are not parallel).
UV_b.x + UV_m.y * t1 = PQ_b.x + PQ_m.x * t2
UV_b.y + UV_m.y * t1 = PQ_b.y + PQ_m.y * t2
Solve the 2D Matrix equation (I assume OP can solve 2D matrix, advise otherwise) If the determinate is 0.0, the lines are parallel, handle that as a special case. (They could be co-indecent and may/may not overlap.)

[UV_m.x - PQ_m.x][t1] = [PQ_b.x - UV_b.x]
[UV_m.y - PQ_m.y][t2] = [PQ_b.y - UV_b.y]
At this point, if t1 and t2 are both in the range 0.0 <= t <= 1.0, the segments intersect! Finding the point of intersection is found simple by Puv(t1) = UV_b + UV_m * t1.

This method handles vertical lines well.

solving two linear equation in program

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