Prove that five points are collinear

  • Thread starter Thread starter Mixer
  • Start date Start date
  • Tags Tags
    Points
AI Thread Summary
To prove that five points are collinear given that no line passes through exactly two points, a contradiction approach is employed. Initially, it is assumed that four points are collinear, with the fifth point being separate. The argument shows that if a line is drawn through any two of the points, it must also include additional points, leading to the conclusion that at least three points must be collinear. The discussion emphasizes the need to address various cases of collinearity among the points to strengthen the proof. Ultimately, the proof is refined by structuring the reasoning and eliminating unnecessary complexities.
Mixer
Messages
38
Reaction score
0

Homework Statement


There are five points on a plane. There is no line that passes trough exactly two points. Prove that five points are collinear.

Homework Equations


-

The Attempt at a Solution


I am just trying to confirm that my proof is correct. I am trying to prove this by contradiction. Let points P1, P2, P3 and P4 be collinear. Point P5 is a separate point. If we define a line that passes trough points P1 and P2 then it also has to pass trough points P3 and P4. Let the slope of that line be k1. If we define another line that passes trough points P1 and P5 for example then it has to have different slope, say k2. That line doesn't pass trough points P2, P3 or P4. Therefore we have defined a line that passes trought exactly two points. That is a contradiction so the five points must be collinear. QED?

I am a little concerned that my proof is not accurate enough. If so what should I add?
 
Physics news on Phys.org
Mixer said:

Homework Statement


There are five points on a plane. There is no line that passes trough exactly two points. Prove that five points are collinear.

Homework Equations


-

The Attempt at a Solution


I am just trying to confirm that my proof is correct. I am trying to prove this by contradiction. Let points P1, P2, P3 and P4 be collinear. Point P5 is a separate point.
Why can you assume this? What if only three are collinear?
If we define a line that passes trough points P1 and P2 then it also has to pass trough points P3 and P4. Let the slope of that line be k1. If we define another line that passes trough points P1 and P5 for example then it has to have different slope, say k2. That line doesn't pass trough points P2, P3 or P4. Therefore we have defined a line that passes trought exactly two points. That is a contradiction so the five points must be collinear. QED?

I am a little concerned that my proof is not accurate enough. If so what should I add?
I think you can proceed directly here. P1 and P2 define a line. There has to be at least one other point on it, say P3. Now you can distinguish the cases P4 is also on this line, and P4 is not. Etc.
 
I am just trying prove this by contradiction. So I am just assuming that point P5 is a separate point. It could be any other point of cource. From there I can come to conclusion that if the points are not collinear then it is possible to define a line that passes trough exactly two points. Which is a contradiction since the assumption was that no such line exists. So do I need something else in my proof or is it done?
 
Mixer said:
I am just trying prove this by contradiction. So I am just assuming that point P5 is a separate point. It could be any other point of cource. From there I can come to conclusion that if the points are not collinear then it is possible to define a line that passes trough exactly two points. Which is a contradiction since the assumption was that no such line exists. So do I need something else in my proof or is it done?
That's the basic idea, yes. But you have to deal with other cases, too. What if only three points are collinear? You cannot automatically assume that four already are.
 
Thank you for your reply!

So my proof is not perfect as I expected. ;-) The case where only three points are collinear seems to be difficult. If points P1, P2 and P3 are collinear then we can define a line that passes trough P1 and P4 or P1 and P5. None of these lines are same as the line that passes trough P1, P2 and P3. Thus a contradiction. What if P1, P4 and P5 are collinear and P1, P2 and P3 are also collinear? Is it sufficient to show we can define a line that passes trough P3 and P5 only?
 
Mixer said:
Thank you for your reply!

So my proof is not perfect as I expected. ;-) The case where only three points are collinear seems to be difficult. If points P1, P2 and P3 are collinear then we can define a line that passes trough P1 and P4 or P1 and P5. None of these lines are same as the line that passes trough P1, P2 and P3. Thus a contradiction.
Contradiction to what?
What if P1, P4 and P5 are collinear and P1, P2 and P3 are also collinear? Is it sufficient to show we can define a line that passes trough P3 and P5 only?
I think you make it more complicate as it is. Stick with your idea:

1.) Two points are always collinear. Say ##s_1=\overline{P_1P_2}##.
2.) By the given condition, there has to be a third point on ##s_1##, say ##P_3##.
3.) If both ##P_4,P_5 \in s_1## we are done. So we may assume that at least one is not, say ##P_4 \notin s_1##.
4.) Then the straight through ##s_2=\overline{P_1P_4}## contains another point, which cannot be ##P_2## or ##P_3##.
5.) Thus ##P_5 \in s_2##.

This is basically what you already wrote in your first post, only structured and with reasons for the assumptions you made.
Now we have the situation ##P_1,P_2,P_3 \in s_1 \neq s_2 \ni P_1,P_4,P_5##.

You don't need a slope here, which is a problematic concept without a coordinate system and not necessary. Just imagine a line ##s_3=\overline{P_5P_3}## and get the contradiction: What is the third point on ##s_3## which is needed for the assumption that no two points are collinear without any others?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top