collinear points..

A

apple2357

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I have been reading through this result and a posting from Dr Math

Desargues' Theorem

What is Desargues' theorem and how can it be proven?
www.nctm.org

Can someone direct me to the proof of a result which is assumed true within this discussion? Quoting from above...

Use the well-known result that if P, Q, and R are collinear, then the
two conditions that will be satisfied are:

k1.p + k2.q + k3.r = 0
k1 + k2 + k3 = 0

where p, q and r are position vectors respectively of P, Q, and R, and
k1, k2, and k3 are scalar constants.

Where does this well-known result come from? Is there a proof someone can help me get started on?
When i think about collinear, i think about the same directions? cant see why the constants will to sum to zero.
 
I'm not familiar with it, but it's easy enough to prove that this is true for any three collinear points.

Points P, Q, and R are collinear when vectors PQ and QR are parallel, i.e. q-p is a scalar multiple of r-q. So suppose that q-p = k(r-q), and see what you can do.

I'll have to see if I can find a "well-known" source for it!

EDIT: The first source I find is Doctor Anthony himself, in another answer (look toward the bottom, where he was asked for a proof): https://www.nctm.org/tmf/library/drmath/view/53382.html
 
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Here is my take on the question when are three points \(P,~ Q,~\&~R\) are collinear?
I recommend A Vector Space Approach to Geometry by Marvin Hausner.
If \(\overrightarrow {PR} = k \cdot \overrightarrow {PQ}~\wedge~k\ne 0\) the three points are collinear.
Moreover, three cases:
\(1)~k>1\text{ then } P-Q-R\\2)~0<k<1\text{ then } P-R-Q\\3)~k<0\text{ then } R-P-Q\)

P.S. About that book. I had a graduate student who used the book as a high school text book and then wrote up the results as his thesis.
 
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Dr.Peterson said:
I'm not familiar with it, but it's easy enough to prove that this is true for any three collinear points.

Points P, Q, and R are collinear when vectors PQ and QR are parallel, i.e. q-p is a scalar multiple of r-q. So suppose that q-p = k(r-q), and see what you can do.

I'll have to see if I can find a "well-known" source for it!

EDIT: The first source I find is Doctor Anthony himself, in another answer (look toward the bottom, where he was asked for a proof): https://www.nctm.org/tmf/library/drmath/view/53382.html
Click to expand...

I don't quite follow Dr anthony's method in the lin but using your idea.. Do i just set up three equations with 3 different constants and combine them somehow?

so we have

q-p=k_1(r-q)

q-p=k_2(r-p)

r-p = k_3(r-q)

Is this what you have in mind or this is an overkill?
 
Start with only what I said, q-p = k(r-q) . Just rearrange this to the form (...)p + (...)q + (...)r = (...), and see what you get.
 
(-1)p+(1+k)q-k r =0



Multiplying through by constant k_1 ?

k_1 p + k_1(1+k)q+kk_1r =0

k_1p + k_2 q +k_3 r =0

where k_1 =-1, k_2 = 0, and k_3= 1

so k_1+k_2+k_3=0


Agree?
 
How are you getting k_2 = 0?

But you don't need to multiply by anything.

You have (-1)p + (1+k)q - k r = 0; now just let k_1 = -1, k_2 = 1+k, and k_3 = -k! What is their sum?
 
Just in case anyone reading doesn't see the following, let me add:

The claim (slightly reworded) was

If P, Q, and R are collinear, then it will be true that:​
​
k1p + k2q + k3r = 0​
k1 + k2 + k3 = 0​
​
where p, q and r are position vectors respectively of P, Q, and R, and k1, k2, and k3 are scalar constants.​

That is, if they are collinear, then there is some such set of constants. We have shown one such set, so we have demonstrated the truth of the statement. In fact, there are infinitely many, since we can just multiply them all by any nonzero number; and we don't need to prove that they are otherwise unique, or to prove the converse (though I imagine both are true).
 
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