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I have 3 lines:

$$ A_1x + B_1y + C_1 = 0 $$ $$ A_2x + B_2y + C_2 = 0 $$ $$ A_3x + B_3y + C_3 = 0 $$

That I believe are dependent. That is, all of the intersections of any pair of two of these lines are exactly the same. Normally I could just show that there is some $\alpha$ such that

$$\alpha(A_1 x + B_1y + C_1) + (A_2x + B_2y + C_2) = (A_3x + B_3y + C_3)$$

but I only have symbolic expressions for the A's, B's, and C's, and they are quite complicated, so finding that $\alpha$ is infeasible. I thought I could instead show that

$$\alpha_A A_1 + A_2 = A_3$$ $$\alpha_B B_1 + B_2 = B_3$$ $$\alpha_C C_1 + C_2 = C_3$$

and show that $\alpha_A = \alpha_B = \alpha_C$, but the $\alpha$'s I find do not seem to be equal (even though plotting the lines shows that they do in fact share a common intersection.

Can anyone comment on what is wrong with this approach? Or suggest a better approach?

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    $\begingroup$ One approach is to solve two of them simultaneously for $x,y$, then plug those values into the third and show satisfaction. $\endgroup$ – Ross Millikan Apr 11 '13 at 20:53
  • $\begingroup$ @RossMillikan - The problem is that the expression of the coefficients are so complicated. When I start diving and substituting things like that, it quickly becomes a big enough mess that I can'st show equivalence. $\endgroup$ – David Doria Apr 12 '13 at 2:20
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The conditions for three distinct lines $A_ix + B_iy + C_i = 0$ in the plane $\mathbb R^2 $ to have a common point of intersection are first that $$ \text {det}\begin {pmatrix} A_1 \:B_1\:C_1 \\ A_2\:B_2\:C_2 \\ A_3\:B_3\:C_3 \\ \end{pmatrix} =0$$ and second that the vectors $ (A_1,A_2,A_3),(B_1,B_2,B_3) \in \mathbb R^3$ be linearly independent.

[The second condition prohibits the three lines from being parallel or, in more sophisticated terms, prohibits their common intersection point from being on the line at infinity of the projective plane $\mathbb P^2(\mathbb R)$]

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  • $\begingroup$ What kind of tolerance would you expect in this determinant numerically? I have these 3 lines: wolframalpha.com/input/… that all intersect at the same point. If I compute the determinant with A1: -1695.48 B1: 1385.93 C1: -3848.26 A2: -27.841 B2: -688.302 C2: 2781.05 A3: 3515.11 B3: -4190.38 C3: 13246.4 determinant = A1*B2*C3 + B1*C2*A3 + C1*A2*B3 - C1*B2*A3 - B1*A2*C3 - A1*C2*B3 = 2048, which seems quite far from zero. $\endgroup$ – David Doria Apr 15 '13 at 14:47
  • $\begingroup$ Dear David, yes 2048 is indeed a big number. I have no explanation for it. $\endgroup$ – Georges Elencwajg Apr 15 '13 at 16:52
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What goes wrong in your approach is the attempt to use only one parameter $\alpha$ to capture the dependence of the third equation on the first two. You need two parameters for this, as the following (EDITED from original answer based on comment by David) shows. Call the equations $E_1,E_2,E_3$ and suppose $$E_1: \ \ x+y=2,\ \ E_2:\ \ x+2y=3.$$ Suppose now that also we have $E_3$ the same as $E_1.$ That is, $$E_3: \ \ x+y=2.$$

Now clearly $E_1,E_2,E_3$ are dependent. If we attempt your method of $\alpha E_1+E_2=E_3$ we will arrive at $$(\alpha+1)x+(\alpha+2)y=2 \alpha+3,$$ and this is not proportional to $E_3$ since for $E_3$ the $x$ and $y$ coefficients are equal.

Basically by using $\alpha E_1 +E_2$ one is committing to using a nonzero multiple of equation $E_2$ in the linear combination of $E_1,E_2$ in the attempt to express (a multiple of) $E_3$, but not all such expressions involve a nonzero contribution from equation $E_2$.

The approach should be able to work if you instead use two parameters $\alpha, \beta$, getting for instance $$\alpha_A A_1 + \beta_A A_2 = A_3,$$ and similarly for the others. It will then be a case of solving the first two equations to determnine the pair $(\alpha, \beta)$ (if that system has a unique solution; if it doesn't the three equations are already dependent or impossible). If all goes well there, you substitute those values of $(\alpha, \beta)$ into the third equation; if it comes out true you have a dependent system, and if false at that point the third equation is not a linear consequence of the first two, making the whole system independent.

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  • $\begingroup$ Hm, I actually started out exactly like that, but said "ah ha, I can just arbitrarily set $\beta = 1$ and then the $\alpha$ that I find will just be adjusted accordingly". That is (in your example), if we had chosen $\frac{3}{4}E_1 + E_2$, would that not be an equivalent expression for $E_3$? Or am I going crazy :) ? $\endgroup$ – David Doria Apr 12 '13 at 2:25
  • $\begingroup$ You're not going crazy. Maybe I was. But if $E_1,E_2$ are as in my example, and it happens that $E_3=E_1$, then there is no $\alpha$ with $\alpha E_1+E_2$ proportional to $E_3$ (which is now $E_1$). I'll change the example in an edit, and include some details. $\endgroup$ – coffeemath Apr 12 '13 at 8:47
  • $\begingroup$ Interesting - you are definitely right about if we only use $\alpha$ we are committing to a non-zero multiple of $E_2$. However, if $E_1 \neq E_3$ then would you agree that we don't need to use a $\beta$? $\endgroup$ – David Doria Apr 12 '13 at 21:44
  • $\begingroup$ I'd say if $E_3$ happens not to be a multiple of $E_1$ you don't need a $\beta$. And whether $E_3$ is a multiple of $E_1$ is an easy check anyway. $\endgroup$ – coffeemath Apr 13 '13 at 2:54
  • $\begingroup$ Consider these lines: wolframalpha.com/input/… $A1 = -1695.48$, $B1 = 1385.93$, $A2 = -27.841$, $B2 = -688.302$, $A3 = 3515.11$, $B3 =-4190.38$ $\alpha_A A_1 + A_2 = A_3$ gives us $\alpha_A = (A_3-A_2)/A_1 = -2.0896$. However, $\alpha_B B_1 + B_2 = B_3$ gives us $\alpha_B = (B_3-B_2)/B_1 = -2.52687$. Can you explain why these are not the same in this case? $\endgroup$ – David Doria Apr 15 '13 at 16:17

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