# Calc. 3 Determining whether points lie in a straight line

by d.tran103
Tags: calc, determining, line, points, straight
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 P: 39 Hey, how do I determine whether or not points lie in a straight line? Is there a symbolic approach to determining so? Or do I need to spatially visualize it? For instance, A(0,-5,5), B(1,-2,4), C(3,4,2) does lie in a straight line according to my book. Thanks!
 P: 772 Do you know any linear algebra? If [0, -5, -5], [1, -2, 4], and [3, 4, 2] all lie in a one dimensional subspace, what does that say about their linear independence?
 P: 367 What is the vector going from A to B (call it v)? If C lies on this line than A-C = tv. Is this true?
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P: 39,682
Calc. 3 Determining whether points lie in a straight line

 Quote by Number Nine Do you know any linear algebra? If [0, -5, -5], [1, -2, 4], and [3, 4, 2] all lie in a one dimensional subspace, what does that say about their linear independence?
Three points can lie on a straight line and NOT lie in a one dimensional subspace so that is NOT sufficient. Only lines that include the origin form a subspace.
P: 4,578
 Quote by d.tran103 Hey, how do I determine whether or not points lie in a straight line? Is there a symbolic approach to determining so? Or do I need to spatially visualize it? For instance, A(0,-5,5), B(1,-2,4), C(3,4,2) does lie in a straight line according to my book. Thanks!
Hey d.tran103 and welcome to the forums.

What you can do is calculate the direction vector of each combination of lines by calculating vectors for lines AB, BC, and AC. Then plug these into a matrix and show whether all of these are linearly dependent.

You can't just plug the raw points in because as HallsOfIvy said, this only works when all lines go through the origin, but if you look at direction vectors of the lines then this is a different thing and you can apply the techniques used in linear algebra.

If they all lie on the same line, then your reduce system will have 1 row of values and the rest will be all 0 elements. If you don't get this situation, then you know that there is at least one other independent characteristic.
 P: 367 The way I think of this is that you consider one of the points as the origin. Then regular linear algebra applies.
P: 4,578
 Quote by deluks917 The way I think of this is that you consider one of the points as the origin. Then regular linear algebra applies.
This is equivalent to what I said above.
 P: 367 ^yes. I am not sure why this matters.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 If you are given the three points to be "A(0,-5,5), B(1,-2,4), C(3,4,2)", then the coordinate system is given and you cannot "take one of them to be the origin". The simplest thing to do is to form the vectors AB= <1- 0, -2- (-5), 4- 5>= <1, 3, 1> and AC== <3- 1, 4- (-2), 2- 4>= <2, 6, -2>. The three points lie on a single line if and only if one of those two vectors is a multiple of the other. Of course, that is equivalent to translating the three points so that A is translated to the origin but that is not just "taking one point to be the origin".

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