In basic geometry we learn about the points, line segments, lines , rays etc. These can be represented on a plane surface which we call it as two-dimensional surface. According to the basic definition, a plane surface contain infinite number of points. Also a line is made up of infinite number of points, which are called as collinear points. By the Euclidean property there exists one and only line passing through two given points. Therefore if we have three points in a plane, there will be two possibilities.
1. It is possible to draw three different lines passing through the three given points.

2. It is possible to draw one and only one line segment passing through the given three given points.

Is there any special name for these points? Yes they are called collinear points. In this section let us see about the definition of collinear points and some examples on Collinear points.

What are Collinear Points ? : Points which lie on the same line are called collinear points.

In the following figure, let us identify the collinear and non-collinear points.
Collinear Points

The points, A, D, O, F and H lie on the same line, Therefore they are called collinear points.

The points B, E, G and I do not lie on the same line, therefore they are called non-collinear points.
Non Collinear Points

In the above figure we observe that the three blue linear intrsect at three points A, B and C.

Also the red line intersects the three linear lines at P, Q and R . Hence we observe that the points P, Q and R are collinear points, and the points A, B and C are non-collinear points.

What are the other collinear points in the above figure?

We can observe that the points, A, R and B ; A, C and P ; B, Q and C are the other sets of collinear points.
If three or more points lie on the same line, then the points are called collinear points.

Let us observe the following figure,
Three Collinear Points
Fig ( 1 ) Fig ( 2 ) Fig ( 3 )


In Fig ( 1 ) we can see that there are infinite number of points in the given plane.
Fig ( 2 ) shows the lines drawn through any two given points.
Fig ( 3 ) represents the lines passing through three points. Therefore the points in Fig ( 3) are collinear points.

Therefore, we have the definition that , if " Three or more points lie on the same line, then the points are called Collinear Points" .

Let us study the following examples :

Solved Example

Question: In the given plane, the lengths of the line segments are as follows.

$\overline{PQ}$ = 7 cm and $\overline{QR}$  = 5 cm. If the points P, Q and R are collinear, find the length of PR.

Line Segments

Solution:
From the above figure, we observe that $\overline{PQ}$ = 7 cm and $\overline{QR}$ = 5 cm

Since the points P, Q and R are collinear, $\overline{PR}$ = $\overline{PQ}$ + $\overline{QR}$

 = 7 + 5

 = 12 cm


Below you could see condition for collinearities
1. If the given points are collinear then the area of the triangle formed by any three points is Zero.

2. The slopes of the lines taken two points at a time are equal.

3. The lines drawn through the points by taking two at a time coincide.

4. If the end points of a line segment is given, then a point is said to be in collinear with the two end points if it lies on the line segment joining the two end points.

Also we have the sum of the two shorter distances is equal to the longest line segment.

If R is a point on the line segment PQ then we have PQ = PR + RQ.

Below you could see example

Example : Show that the points ( -5, 7 ), ( - 4, 5 ) and ( 1, -5 ) are collinear.


Method 1 : Using Area Formula :

Solution : If the points are collinear, the area of the triangle formed by the three points will be 0.
We have the area formula, Area = $\frac{1}{2}$ { x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 ) }

Substituting the above points in the formula, we get,

Area = $\frac{1}{2}$ { -5 ( 5 + 5 ) -4 ( -5-7) + 1 ( 7 - 5 ) }

= $\frac{1}{2}$ { ( -5 ) ( 10 ) -4 ( -12 ) + 1 ( 2 ) }

= $\frac{1}{2}$ { -50 + 48 + 2 }

= $\frac{1}{2}$ { -50 + 50}

= $\frac{1}{2}$ ( 0 )

= 0

Since the area of the triangle formed by the three points is equal to 0, the three points are collinear.

Method 2 : Using Distance formula:

Solution: Let the three points be, A, B and C respectively.
We have the distance formula between the two points ( x1 , y1 ) and ( x2 , y2 ) given by,
d = $\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
We have A ( -5, 7 ), B( -4, 5) and C(1,-5)

Therefore, we have AB = $\sqrt{(-4+5)^{2}+(5-7)^{2}}$
= $\sqrt{1+4}$
= $\sqrt{5}$

BC = $\sqrt{(1+4)^{2}+(-5-5)^{2}}$
= $\sqrt{25+100}$
= $\sqrt{125}$
= 5 $\sqrt{5}$

CA = $\sqrt{(1+5)^{2}+(-5-7)^{2}}$
= $\sqrt{36+144}$
= $\sqrt{180}$
= 6 $\sqrt{5}$

From the lengths of the sides, we observe that,
$\sqrt{5}$ + 5 $\sqrt{5}$ = 6 $\sqrt{5}$
=> AB + BC = AC
Therefore, the three points, A, B and C are collinear.

Method 3 : Using slopes

Solution: When the points are collinear, the slopes between pair of points among the three given points will be equal.
The slope of the line joining the points ( x1 , y1 ) and ( x2 , y2 ) is $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
We have, A ( -5, 7 ), B( -4, 5) and C(1,-5)
Using the above formula,
Slope of AB = $\frac{(5-7)}{(-4+5)}$
= $\frac{-2}{1}$
= - 2
Slope of BC = $\frac{-5-5}{1+4}$
= $\frac{-10}{5}$
= - 2
Slope of AC = $\frac{-5-7}{1+5}$
= $\frac{-12}{6}$
= - 2

We observe that Slope of AB = Slope of BC = Slope of AC

=> The three points are collinear.