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So lets say I have some points $A,B,C$.

A method I have been shown for determining if the lie along a straight line is thus:
$\mathrm{If}\space|AC|=|AB|+|BC| \space\mathrm{then\space A,B\space and\space C\space lie\space on\space a\space straight\space line. }$

However, this doesn't seem like a very reliable method to me. It seems like a situation such as this would create a false positive. enter image description here

If this is not the case, how do you prove that $|AB|+|BC|\neq|AC|$ in all cases if they points are not on a straight line?

If this case will indeed cause the algorithm to fail, what reliable algorithm can I use?

Thanks...

EDIT:: Whenever I built a triangle that would cause this algorithm to fail, (ie, a=10,b=5,c=5 and others) I get $cos{C}=1$, implying a straight, line? How can one prove this is always the case?

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user1833028
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  • Why should the equality hold in the second example ? – Peter May 06 '14 at 17:10
  • It's a poorly drawn image, I know. But, can the points form a triangle where one arm is twice as long as the other, thus an arbitrarily chosen point A could possibly be twice as far away from point C as it is from point B, thus,$|AB|+|BC|=|AC|+2|AB|=2|BC|$ – user1833028 May 06 '14 at 17:17
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    Colinearity is usually checked using cross product. You might find this useful. – Marcin Łoś May 06 '14 at 17:17
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    But the claim |AB| + |BC| = |AC| does not hold, if A,B and C form a triangle. – Peter May 06 '14 at 17:19
  • I've tried to construct several such triangles and they always end up with a 0 degree angle! But how can I prove this is ALWAYS the case? – user1833028 May 06 '14 at 17:31

2 Answers2

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The equality $|AB| + |BC| = |AC|$ holds if and only if point $B$ lies on the segment $AC$.

Still, this is a bad basis for a practical algorithm. If you implement this, you'll have to calculate three square roots, which is slow. Not to mention that this just looks bad.

A good way to determine whether or not points $A$, $B$ and $C$ are collinear is purely linear. It goes like this.

Calculate coordinates of vectors $\vec{AB}$ and $\vec{BC}$: $\vec{AB} = (x_1, y_1, z_1)$ and $\vec{AC} = (x_2, y_2, z_2)$. Now, points $A$, $B$ and $C$ are collinear if and only if all the equalities below are true: $$ \begin{align*} x_1 y_2 &= x_2 y_1 \\ y_1 z_2 &= y_2 z_1 \\ z_1 x_2 &= z_2 x_1 \end{align*} $$

Dan Shved
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A little extension makes the algorithm work :

Consider $|AB|$ , $|AC|$ and $|BC|$. Then, C is on the line AB , if and only if the greatest of the three numbers is the sum of the others. In Particular, if C = A or C = B, then C is trivially always on the line AB.

Peter
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