Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
13 May 2013, 08:47

1

This post received KUDOS

Narenn wrote:

Question Source :- My own question

On the x y plane, there are 8 points of which 4 are collinear. How many straight lines can be formed by joining any 2 points from the 8 points ?

A) 28 B) 22 C) 20 D) 23 E) 56

OA and OE after some discussion. +1 Kudo for Each correct and Detailed explanation

Regards

Narenn

I believe the answer should be 23 lines i.e [D]

The approach taken by me would be simple counting: 1. We can always say that one line is to pass though the collinear points, let the names be A, B, C and D. 2. The other four points are scattered under the x-y plane. Hence Each point can form 4 lines with the A,B,C and D. Hence, the total lines would be 16. 3. Finally, the total number of lines between the 4 non-collinear points would be the same as the 4C2 i.e 6. I am assuming here the rest of points are non-collinear and no line can also be drawn among any 3 of them.

Hence the total comes to be 23. Hope I am correct!

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
13 May 2013, 08:53

1

This post received KUDOS

Each line is defined by two points.

4 points are collinear => 1 line passes through those. Now each of those points form 5 line => 4 lines with the "out" points and 1 line is the common one but this is counted (the common) so we just add 4(point)*4(lines)=16 to the sum. Finally the 4 points "out" of the common line form 4C2 = 6 different lines

1+16+6=23 totals
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
13 May 2013, 10:23

My answer is A 28.(quite skeptical).

Let the 4 collinear points be A,B,C,D. then I can join AB as 1 line, BC as 1 line,CD as 1 line. AC as 1 line. AD as 1, BC,CD as 2 more. Giving a total of 6 lines.--(1) My reasoning here is question says the number of line that CAN be formed by joining any 2 points. So all the above lines can be treated as different lines between 2 points. Like line AD passes through points B and C but in essence it is a different line to BC or CD.

Next each non collinear point will have 1 passing through A,B,C and D..hence all 4 non collinear pints will in total have 16 lines.--(2)

And between the 4 non collinear we can for 6 lines.(4C2) or you can visualise these 4 points to be the corners of a square so you will have 4 sides and 2 diagonals. total 6 lines.--(3)

hence total number of lines are (1)+(2)+(3) = 6+16+6=28.

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
13 May 2013, 10:27

mdbharadwaj wrote:

My answer is A 28.(quite skeptical).

Let the 4 collinear points be A,B,C,D. then I can join AB as 1 line, BC as 1 line,CD as 1 line. AC as 1 line. AD as 1, BC,CD as 2 more. Giving a total of 6 lines.--(1) My reasoning here is question says the number of line that CAN be formed by joining any 2 points. So all the above lines can be treated as different lines between 2 points. Like line AD passes through points B and C but in essence it is a different line to BC or CD.

Next each non collinear point will have 1 passing through A,B,C and D..hence all 4 non collinear pints will in total have 16 lines.--(2)

And between the 4 non collinear we can for 6 lines.(4C2) or you can visualise these 4 points to be the corners of a square so you will have 4 sides and 2 diagonals. total 6 lines.--(3)

hence total number of lines are (1)+(2)+(3) = 6+16+6=28.

Umm mdbharadwaj don't you think the 6 line-segments joining the collinear points can actually considered to be one unique line. If we are not considering unique lines, then the number is bound to increase significantly.

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
13 May 2013, 10:41

1

This post received KUDOS

arpanpatnaik wrote:

mdbharadwaj wrote:

My answer is A 28.(quite skeptical).

Let the 4 collinear points be A,B,C,D. then I can join AB as 1 line, BC as 1 line,CD as 1 line. AC as 1 line. AD as 1, BC,CD as 2 more. Giving a total of 6 lines.--(1) My reasoning here is question says the number of line that CAN be formed by joining any 2 points. So all the above lines can be treated as different lines between 2 points. Like line AD passes through points B and C but in essence it is a different line to BC or CD.

Next each non collinear point will have 1 passing through A,B,C and D..hence all 4 non collinear pints will in total have 16 lines.--(2)

And between the 4 non collinear we can for 6 lines.(4C2) or you can visualise these 4 points to be the corners of a square so you will have 4 sides and 2 diagonals. total 6 lines.--(3)

hence total number of lines are (1)+(2)+(3) = 6+16+6=28.

Umm mdbharadwaj don't you think the 6 line-segments joining the collinear points can actually considered to be one unique line. If we are not considering unique lines, then the number is bound to increase significantly.

Regards, Arpan

I thought of that, however the questions asks for the number of line segments that can be formed between any 2 points. Consider numbers 1,2,3,4 on the number line. So line 1-2 is different from line 3-4. 1-2 is an unique line between points 1 and 2 so is 3-4. And the number doesn't increase significantly , as there will be a maximum of 28 lines. Because if we consider the line joining the 4 collinear points to be 1 single line or to be 6 different lines, the number of lines between the the non collinear points themselves and the collinear points will not change. there will be 16+6 = 22 lines only.

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
13 May 2013, 11:00

1

This post received KUDOS

mdbharadwaj wrote:

I thought of that, however the questions asks for the number of line segments that can be formed between any 2 points. Consider numbers 1,2,3,4 on the number line. So line 1-2 is different from line 3-4. 1-2 is an unique line between points 1 and 2 so is 3-4. And the number doesn't increase significantly , as there will be a maximum of 28 lines. Because if we consider the line joining the 4 collinear points to be 1 single line or to be 6 different lines, the number of lines between the the non collinear points themselves and the collinear points will not change. there will be 16+6 = 22 lines only.

Hope I got my point across.

Lemme just clarify my idea over this. Please refer to my *supremely sloppy* paint-work attached (Apologize for that ). You can see, that A, B and C lie on the line L. Now As the question states:

Quote:

How many straight lines can be formed by joining any 2 points from the 8 points ?

Now, If I am to consider a straight line passing through A and B, it would be L. Again if I am to consider a straight line passing through B and C, it would be L. Same is the case for A and C as well. The A-B, B-C and C-A are segments. L is the only unique straight line passing through the collinear points. Getting back to the question, you are spot-on for the 16 and 6 values. After adding the one line that passes through the collinear points, I believe you have the answer! Hope I am correct! Mods please verify!

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
13 May 2013, 15:48

2

This post received KUDOS

Narenn wrote:

Question Source :- My own question

On the x y plane, there are 8 points of which 4 are collinear. How many straight lines can be formed by joining any 2 points from the 8 points ?

A) 28 B) 22 C) 20 D) 23 E) 56

OA and OE after some discussion. +1 Kudo for Each correct and Detailed explanation

Regards

Narenn

Number of ways two points can be selected from the 8 points is 8C2. Number of lines that can be formed by the collinear points if they were non collinear is 4C2. Therefore total number of lines is 8C2 - 4C2 + 1(one for the line which is formed by the collinear points)
_________________

"Kudos" will help me a lot!!!!!!Please donate some!!!

Completed Official Quant Review OG - Quant

In Progress Official Verbal Review OG 13th ed MGMAT IR AWA Structure

Yet to do 100 700+ SC questions MR Verbal MR Quant

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
14 May 2013, 11:16

Expert's post

Dear Responders,

Thank you all for participating in this quiz, coming with your detailed solutions, and having a thoughtful debate here. I would also like to congratulate all interns as well for being the part of this glorious community.

At first I would like to inform you that this question and its logic is based on the question given in Indian school text book (CBSE / Class XI / Volume II / Chapter 35 - Combinations / Page 35.13 / Author - Shri R. D. Sharma). The snapshot of this reference question is also given herewith for your clear understanding.

We have 8 points on x y plane of which 4 are collinear. We should know that the 4 collinear points can not form any other line among them except for 1 that will connect two extremes. (We are discussing here about a line and not about a line segment)

Connecting any two points from 8 points is similar to choose 2 things from 8. This can be done in 8C2 ways. Note these 8C2 combinations also include the combination of fictitious lines that can be formed from 4 collinear points. This we can calculate as 4C2 We also have to include 1 line (formed by connection collinear extremes) in above combinations.

Hence Total Number of Combinations will be 8C2 - 4C2 + 1 ------> 28 - 6 + 1 = 23 = Choice D.

This logic we can take further in triangle case Number of triangles = 8C3 - 4C3 (Here collinear extremes will not form any triangle, so no need to add 1 in this case)

mdbharadwaj wrote:

I thought of that, however the questions asks for the number of line segments that can be formed between any 2 points.

Question indeed asks for number of lines and not for number of line segments.

Those who gave correct solution have been awarded Kudos. In exceptional case, Kudos has also been awarded to 'mdbharadwaj' for his honest fight and to encourage him/her for further contribution. Thank You,

Regards,

Narenn

Attachments

CBSE P&C.png [ 286.58 KiB | Viewed 901 times ]

lines combinations.png [ 25.48 KiB | Viewed 897 times ]

Re: On the x y plane, there are 8 points of which 4 are collinea [#permalink]
14 May 2013, 12:00

Narenn wrote:

Question Source :- My own question

On the x y plane, there are 8 points of which 4 are collinear. How many straight lines can be formed by joining any 2 points from the 8 points ?

A) 28 B) 22 C) 20 D) 23 E) 56

OA and OE after some discussion. +1 Kudo for Each correct and Detailed explanation

Regards

Narenn

8 points total - 4 are collinear (lie along a straight line) and 4 are non-collinear.

a) All the collinear points lie on a straight line - 1 line can be drawn through them. b) Use combinations for the remaining 4 non-collinear points. We use combinations because, the order of the points along a line do not matter. We have 4 total points to choose from, and we can choose only 2 points to draw a line through. Hence, it is 4C2. 4C2 = 4!/(2!*2!) = 6. Thus, 6 lines can be drawn here. c) Each non-collinear point can draw a unique line through each collinear point. That is, 4 non-collinear points * 4 collinear points = 16 possible lines.

Thus there are 1+6+16=23 possible lines that can be drawn. The correct answer should be D.