# Collinear Points

• Sep 15th 2007, 05:37 AM
sk8ingkittty
Collinear Points

Definition: Points that lie on the same line are called collinear.
Not sure if you need a minimum of three points in on a same line to be considered collinear.

Question

Five point A-E, NO definite lines drawn. E, D and C appear to be in line, A is below D, B is below C

Conside points A,B,C,D and E as shown

1. If two of these points are selected at random, what is the probability that they are collinear?
2. If three of these points are selected at random, what is the probability that they are collinear?
3. If four of these points are selected at random, what is the probability that they are collinear?
• Sep 15th 2007, 05:47 AM
Plato
Without actually seeing the diagram there is no way to answer parts (b) & (c).
However, the answer to (a) is 1: any two points determine a line.

Can you use a PAINT type program to insert the diagram?
• Sep 15th 2007, 07:26 AM
liz155
http://img.photobucket.com/albums/v2.../1geometry.jpg

Here's the picture. I am a friend of sk8ingkittty and I also tried to figure this out but it seems there is not enough info to do that. Thanks for your help.
• Sep 15th 2007, 07:56 AM
Plato
O.K. then.
I thought that might be the diagram. I assume that eab are meant to be in a line.

The combination of five points taken three at a time, $\displaystyle \left( {\begin{array}{c} 5 \\ 3 \\ \end{array}} \right) = 10$.
From the diagram there are only two subsets of three that are collinear. So what is the answer to (b)?

$\displaystyle \left( {\begin{array}{c} 5 \\ 4 \\ \end{array}} \right) = 5$, how many subsets of four are collinear? So what is the answer to (C)?
• Sep 15th 2007, 08:20 AM
Soroban
Hello, sk8ingkittty!

From your sketch, I will assume that the layout looks like this:
Code:

      E      D      C       *      *      *               *               A                       *                       B
I further assume that $\displaystyle E,D,C$ are collinear
. . and that $\displaystyle E,A,B$ are collinear.

Quote:

Consider points $\displaystyle A,B,C,D,E$ as shown.

1. If two of these points are selected at random,
what is the probability that they are collinear?

There are: .$\displaystyle {5\choose2} \:=\:10$ pairs of points.

Any two points are collinear.
. . Hence, there are $\displaystyle 10$ pairs of collinear points.

Therefore: .$\displaystyle P(\text{2 points collinear}) \:=\:\frac{10}{10}\:=\:1$

Quote:

2. If three of these points are selected at random,
what is the probability that they are collinear?

There are: .$\displaystyle {5\choose3}\,=\,10$ sets of three points.

There are 2 sets which are collinear: .$\displaystyle EDC,\:EAB$

Therefore: .$\displaystyle P(\text{3 points collinear}) \:=\:\frac{2}{10} \:=\:\frac{1}{5}$

Quote:

3. If four of these points are selected at random,
what is the probability that they are collinear?

There are: .$\displaystyle {5\choose4} \,=\,5$ sets of four points.

There are no sets of four points that are collinear.

Therefore: .$\displaystyle P(\text{4 points collinear}) \:=\:\frac{0}{5} \:=\:0$

• Sep 18th 2007, 02:45 PM
sk8ingkittty
Thank You for the help. I'm still confused on how do you find out how many pairs of points there are.