P&c q4

[Saumyojit]

there are 4 straight lines in a plane so that no two lines are parallel and no three lines are concurrent determine the number of points of intersection .

I know what concurrent means that three lines or more cannot have a common intersection point.


Now before any approach I wanted to draw the Diagram.
Firstly, i draw the Diagram in such a way where every line was cutting every other except the last 4 th line which cut only two lines.
So the no of intersection points came 5 .
But answer is 6 .

Then I draw the Diagram where every line cuts each other then came 6 Intersection points.(4c2)

So why the question demands that every line should cut each other?

 

[Dr.Peterson]

So why the question demands that every line should cut each other?

What is the definition of "parallel"? And how could you draw a diagram in which no lines are parallel but one cuts only two of the others?

I suspect that you drew segments, not lines which extend infinitely (and therefore missed an intersection).

(Thanks for telling us this time what the answer is; but why not show us your pictures?)

 

[pka]

there are 4 straight lines in a plane so that no two lines are parallel and no three lines are concurrent determine the number of points of intersection . I know what concurrent means that three lines or more cannot have a common intersection point. Now before any approach I wanted to draw the Diagram.
Firstly, i draw the Diagram in such a way where every line was cutting every other except the last 4 th line which cut only two lines.
So the no of intersection points came 5 . But answer is 6 .

Others may disagree, but in this case I would advise against trying a diagram. I would advise you to think about the logic involved with four lines no two of which are parallel and no three of which are concurrent.
Now \(^4\mathcal{C}_2=\dbinom{4}{2}=\dfrac{4!}{2!\cdot 2!}=6\). How do we know that the six points are all distinct?

 

[Saumyojit]

I suspect that you drew segments, not lines which extend infinitely (and therefore missed an intersection).

If 2 non parallel lines are drawn in a plane, it is certain that they will intersect somewhere

i got my answer .

 

[Saumyojit]

Find the number of straight lines obtained by joining N points on a plane, if No three of which are collinear
What will be the number if p points out of n are in same strt line?

First one , gives me nC2 then the next question .....
Meaning of last line ? if n=8 ,p=4 then what is the diagram ?

 

[Dr.Peterson]

Find the number of straight lines obtained by joining N points on a plane, if No three of which are collinear
What will be the number if p points out of n are in same strt line?

First one , gives me nC2 then the next question .....
Meaning of last line ? if n=8 ,p=4 then what is the diagram ?

Once again, can you show us the source, so we can check the context and wording, since you clearly did not paste it in (at least, not from a good source)? Is there an answer provided?

I would draw a line and put 4 points on it, then add 4 more points, none of them collinear.

Or rather, I would first imagine doing so and try solving the problem without drawing, and then draw to check my answer. This would also help me to check my interpretation of the problem; if what I do doesn't make sense, I must be wrong. But I think I'm right.

 

[Saumyojit]

can you show us the source

the question is given in the book. I have wrote it correctly .


Find the number of straight lines obtained by joining N points on a plane, if No three of which are collinear.
What will be the number of straight lines if p points out of n are in same straight line?

Is there an answer provided?

yes, in a expression .

I would draw a line and put 4 points on it, then add 4 more points, none of them collinear.

Yes i thought like this only
4 points are on the same line .
Then another 4 points are divided into 2 pairs .

A ----B----C----D

E----F

G---H

Here st line will be abcd , ef, gh, ae, eg, af etc .... like this right?

The above diagram can it be like this

A ----B----C----D
E----F---G
.H

If p=4 then 4 points can be colinear then three points can be colinear also right? Any no of points less than p

 

[Dr.Peterson]

the question is given in the book. I have wrote it correctly .

No you have not. The spelling and grammar both make it clear that you (or they?) were not careful. (And the right word is "written"; I won't hold that against you.)

Find the number of straight lines obtained by joining N points on a plane, if No three of which are collinear.
What will be the number of straight lines if p points out of n are in same straight line?

Surely it should have been "no three of which are collinear". At least this time you corrected "strt".

yes, in a expression .

So what IS it? Haven't you learned from the previous discussion??

Yes i thought like this only
4 points are on the same line .
Then another 4 points are divided into 2 pairs .

A ----B----C----D

E----F

G---H

Here st line will be abcd , ef, gh, ae, eg, af etc .... like this right?

The above diagram can it be like this

A ----B----C----D
E----F---G
.H

If p=4 then 4 points can be colinear then three points can be colinear also right? Any no of points less than p

There is no need to divide the four into pairs; and if that number were even, you couldn't. Don't add to what the problem says.

You have 4 points on a line, and 4 other points, no three of which are collinear. I would have preferred for the problem to explicitly say that, but I think it is implied by the fact that if we allowed some other points to be collinear besides the 4, then the problem couldn't be solved. Additional collinear sets change the answer.

Just think about how you can form lines using the points you have. As was said before, although making a diagram can help to raise questions about the meaning of the problem, you should think generally rather than focus on one diagram.

 

[Saumyojit]

right word is "written

yes. i have written .Thanks

if that number were even, you couldn't

did not understand

You have 4 points on a line, and 4 other points, no three of which are collinear

yes, in the second question it's not explicitly mentioned .
So, it will be like this

pt ----pt----pt----pt

pt----pt

pt---pt

4 other points, no three of which are collinear

I need to select which are those p points out of n which will be on the same st line . ?

I think it is implied by the fact that if we allowed some other points to be collinear besides the 4, then the problem couldn't be solved

why?

So what IS it?

nC2 - pC2 + 1.

 

[pka]

did not understand
yes, in the second question it's not explicitly mentioned . So, it will be like this
pt ----pt----pt----pt
pt----pt
pt---pt
4 other points, no three of which are collinear
I need to select which are those p points out of n which will be on the same st line . ?
nC2 - pC2 + 1.

Did you read reply #3?

 

[Saumyojit]

Did you read reply #3?

see#4.

 

[Dr.Peterson]

nC2 - pC2 + 1.

I suppose your difficulty is not being sure how to interpret the problem. It would help a lot if you would just show an attempt at the solution (explaining what you do), which would reveal your thinking better.

What kinds of lines are there to count? There is the one line containing the given p points; then there are the lines determined by each of the p points together with each of the other (n-p) points; and then there are the lines formed by two of the (n-p) points.

Count them and add.

Or, as their answer suggests, you might try a subtractive method, which should give an equivalent expression. This might involve starting with the total number of lines ignoring the known collinearity, and subtracting lines that are counted more than once. This will be easier.

why?

I answered that in the sentence after the one you quoted! The answer would depend on how many sets of points are collinear, so it would not be unique.

 

[Saumyojit]

There is the one line containing the given p points; then there are the lines determined by each of the p points together with each of the other (n-p) points; and then there are the lines formed by two of the (n-p) points.

Count them and add.

yes .
It will be nC2- pc2 +1.
I should have done this .
Shame on Me.

 

[Steven G]

Let the name of the four lines be 1, 2, 3 and 4
1 and 2 intersect.
1 and 3 intersect.
1 and 4 intersect.
2 and 3 intersect.
2 and 4 intersect
3 and 4 intersect

 

[Dr.Peterson]

Let the name of the four lines be 1, 2, 3 and 4
1 and 2 intersect.
1 and 3 intersect.
1 and 4 intersect.
2 and 3 intersect.
2 and 4 intersect
3 and 4 intersect

What 4 lines are you talking about? I don't see any discussion here of four lines.

yes .
It will be nC2- pc2 +1.
I should have done this .
Shame on Me.

You said this in reply to my suggestion of a different method, using addition! Did you really try that method?

There is the one line containing the given p points; then there are the lines determined by each of the p points together with each of the other (n-p) points; and then there are the lines formed by two of the (n-p) points.

For n=8, p=4, this gives 1, plus 4*4 =16, plus 4C2 = 6, for a total of 23; the formula obtained the other way gives 8C2-4C2+1 = 28-6+1 = 23. Proving the two formulas equivalent is harder.

 

[Steven G]

What 4 lines are you talking about? I don't see any discussion here of four lines.

@Dr.Peterson
there are 4 straight lines in a plane so that no two lines are parallel and no three lines are concurrent determine the number of points of intersection

 

[Dr.Peterson]

@Dr.Peterson
there are 4 straight lines in a plane so that no two lines are parallel and no three lines are concurrent determine the number of points of intersection

You're right; but the current discussion was of a new question in #5.