Collinearity
As per the Euclidean geometry, a set of points are considered to be collinear, if they all lie in the same line, irrespective of whether they are far apart, close together, form a ray, a line, or a line segment. The term 'collinear' is derived from a Latin word where 'col' means 'together' and linear means 'line'. Thus, collinear, means the points lying together on a single line. Collinear points can be seen in our daytoday lives, for example, the eggs placed in a row, or, the numbers printed on a ruler are collinear.
1.  Collinearity 
2.  How To Determine if the Points Are Collinear? 
3.  Solved Examples 
4.  Practice Questions 
5.  FAQs on Collinearity 
Collinearity
As per the collinearity property, three or more than three points are said to be collinear when they all lie on a single line. The term collinearity has been used to indicate that the objects lie on a single line or in a row. Let us have a look at an example to understand collinearity better.
Example:
Consider the following three points: A (−3,−1), B (−1,0), and C (1,1). If we plot these points in the cartesian plane, we find that they are collinear.
Collinear and NonCollinear Points
When a line passes through three or more points, the points are said to be collinear points. In other words, the points lying on a straight line are called collinear points. In the following figure, the points C, B, and A are collinear as they all lie on the line 'q'. The points E, B, and D are also collinear as they lie on the line 'p'. If it is not possible to draw a straight line through three or more points, then they are said to be noncollinear points. Points D, B, and A are noncollinear points since there is no line that passes through all these three points.
How To Determine if the Points Are Collinear?
There are various methods to find out whether three points are collinear or not. These methods are given as follows:
 Using Distance Formula
 Slope Method
 Area of Triangle Method
Using Distance Formula
Consider three points P, Q, and R. Three points P, Q, R will be collinear if: length of PQ + length of QR = length of PR
The length of the given points can be calculated with the help of the distance formula.
\(d=\sqrt{\left(x_{2}x_{1}\right)^{2}+\left(y_{2}y_{1}\right)^{2}}\)
Example: Let us find whether the three points A (−3,−1), B (−1,0) and C (1,1) are collinear or not.
Solution: Three points will be collinear if AB + BC = CA
\( \begin{align*} \text{Length of AB} &= \sqrt{\left(x_{2}x_{1}\right)^{2}+\left(y_{2}y_{1}\right)^{2}} \\ &= \sqrt{(1+3)^{2}+(0+1)^{2}} \\ &= \sqrt{5} \end{align*}\)
\( \begin{align*} \text{Length of BC} &= \sqrt{\left(x_{3}x_{2}\right)^{2}+\left(y_{3}y_{2}\right)^{2}} \\ &=\sqrt{(1+1)^{2}+(10)^{2}} \\ &= \sqrt{5} \end{align*}\)
\( \begin{align*} \text{Length of AC} &= \sqrt{\left(x_{3}x_{1}\right)^{2}+\left(y_{3}y_{1}\right)^{2}} \\ &= \sqrt{(1+3)^{2}+(1+1)^{2}} \\ &= \sqrt{20} = \sqrt{(4 \times 5)} = 2\sqrt{5} \end{align*}\)
This means AB + BC = CA because √5 + √5 = 2 √5. Therefore, it is proved that the three points A, B and C are collinear.
Slope Method
As per the slope method: If three points P(\(x_1\),\(y_1\)),Q(\(x_2\),\(y_2\)),R(\(x_3\),\(y_3\)) are collinear, then the slope of the line formed by points P(\(x_1\),\(y_1\)),Q(\(x_2\),\(y_2\)) and Q(\(x_2\),\(y_2\)),R(\(x_3\),\(y_3\)) will be equal.
The slope of line QR is:\(m_{1}=\dfrac{y_{3}y_{2}}{x_{3}x_{2}}\)
The slope of line PQ is:
\(m_{2}=\dfrac{y_{2}y_{1}}{x_{2}x_{1}}\)
\(m_{1}=m_{2}\)
\(\begin{equation}
\dfrac{y_{3}y_{2}}{x_{3}x_{2}}=\dfrac{y_{2}y_{1}}{x_{2}x_{1}}
\end{equation}\)
Example: Let us find out whether the points P, Q and R are collinear or not using the slope method.
Solution:In this case, let us assume (x_{1} , y_{1}) = (1, 2), (x_{2} , y_{2}) = (2, 3), (x_{3} , y_{3}) = (3, 4)
Now, let us find the slope of line QR after placing the values in the equation:
\(\begin{align*} m_{1} &=\dfrac{y_{3}y_{2}}{x_{3}x_{2}} \\ m_{1} &=\dfrac{43}{32} \\ m_{1} &=\dfrac{1}{1} \\ m_{1} &=1 \end{align*}\)
Slope of line PQ is:\(\begin{align*} m_{2} &=\dfrac{y_{2}y_{1}}{x_{2}x_{1}} \\ m_{2} &=\dfrac{32}{21} \\ m_{2} &=\dfrac{1}{1} \\ m_{2} &=1 \\ m_{1} &=m_{2} \end{align*}\)
Therefore, the given points are collinear because m_{1} = m_{2}
Area of Triangle Method
As per the area of triangle method, if three points P(\(x_1\),\(y_1\)),Q(\(x_2\),\(y_2\)),R(\(x_3\),\(y_3\)) are collinear, the area of the triangle formed by all three points P(\(x_1\),\(y_1\)),Q(\(x_2\),\(y_2\)), and R(\(x_3\),\(y_3\)) will be zero.
If three points are collinear, the area of a triangle formed by the three points on a cartesian plane is zero. The formula for the area of a triangle with the coordinates is written as:
\( \text{A} =\frac{1}{2}\left\left(x_{1}\left(y_{2}y_{3}\right)+x_{2}\left(y_{3}y_{1}\right)+x_{3}\left(y_{1}y_{2}\right)\right)\right=0\)
Example: Check whether the points (−1,−1),(1,1) and (3,3) are collinear or not.Solution: Let's take (−1,−1) as (\(x_1\),\(y_1\)),(1,1) as (\(x_2\),\(y_2\)), and (3,3) as (\(x_3\),\(y_3\)) .
Substituting these value in the formula of area of the triangle:
\(\text{A} =\frac{1}{2}\left\left(x_{1}\left(y_{2}y_{3}\right)+x_{2}\left(y_{3}y_{1}\right)+x_3\left(y_{1}y_{2}\right)\right)\right=0 \\ A=\frac{1}{2} \mid(1(13))+1(3(1))+3(11) \mid=0 \\ A=\frac{1}{2}(2+46)=0 \\ Area=\frac{1}{2}0=0 \\ Area=0 \)
Hence, the points are collinear.
Related Topics
Check out these interesting articles to learn more about collinearity and its related topics
 Slope
 X and Y Graph
 Coordinate Plane
 Equation of a Straight Line
 Area of triangle in coordinate geometry
 Distance between Two Points
 Equation of a Line
Important Notes
Here is a list of a few points that should be remembered while studying collinearity:
 As per the collinearity property, three or more than three points are said to be collinear when they all lie on a single line.
 When a line passes through three or more points, the points are said to be collinear points.
 If it is not possible to draw a straight line through three or more points, then they are said to be noncollinear points.
Solved Examples

Example 1: Prove that the points P(4, 2), Q(0, 2), and R(3, 5) are collinear points.
Solution:
Let's prove that the points are collinear using the slope method. If three points are collinear, they will satisfy the following condition:
\(\dfrac{y_{3}y_{2}}{x_{3}x_{2}}=\dfrac{y_{2}y_{1}}{x_{2}x_{1}}\)
Here,
(\(x_1\),\(y_1\)) is (−4,−2)
(\(x_2\),\(y_2\)) is (0,2)
(\(x_3\),\(y_3\)) is (3,5)Put the values in the above equation,
\(\dfrac{y_{3}y_{2}}{x_{3}x_{2}} = \dfrac{y_{2}y_{1}}{x_{2}x_{1}} \\ \dfrac{52}{30} = \dfrac{2(2)}{0(4)} \\ \dfrac{3}{3} =\dfrac{4}{4} \\ 1 =1 \\ \text{ L.H.S. } = \text{ R.H.S. }\)
Hence, the points are collinear.
Answer: Points P(4, 2), Q(0, 2) and R(3, 5) are collinear.

Example 2: Find the value of k for which the given points J(0, 2), K(2, 0), and L(5, k) are collinear.
Solution:
Given: Points J, K, and L are collinear.
If three points are collinear, the area of the triangle formed by these points will be zero.
Here,
(\(x_1\),\(y_1\)) is (0,−2)
(\(x_2\),\(y_2\)) is (2,0)
(\(x_3\),\(y_3\)) is (5,k)
\(\text {Area of triangle }=\frac{1}{2} \mid\left(x_{1}\left(y_{2}y_{3}\right)+x_{2}\left(y_{3}y_{1}\right)+x_{3}\left(y_{1}y_{2}\right) \mid=0\right.\)Substituting the values in the above equation, we get,
\(\dfrac{1}{2} \mid(0(0k)+2(k(2))+5(20) \mid=0 \\ \dfrac{1}{2}(0+2 k+410) =0 \\ \dfrac{1}{2}(2 k6) =0 \\ (k3) =0 \\ k =3 \)
Answer: Therefore, the value of 'k' is 3.
FAQs on Collinearity
What is Collinearity in Geometry?
In geometry, three or more points are considered to be collinear if they all lie on a single straight line. This property of the points is called collinearity.
What is the Formula of Collinearity?
If P, Q, and R are three collinear points, then, PQ + QR = PR. Another formula that proves three points to be collinear is the area of a triangle formula. After substituting the coordinates of the given points in the formula, if the value is equal to zero, then the given points will be collinear. For example, if three points A (a_{1}, b_{1}), B (a_{2}, b_{2}) and C (a_{3}, b_{3}) are collinear, then [a_{1}(b_{2}  b_{3}) + a_{2}( b_{3}  b_{1})+ a_{3}(b_{1}  b_{2})] = 0.
How Do You Prove Collinearity?
As per the Euclidean geometry, a set of points are considered to be collinear, if they all lie in the same line, irrespective of whether they are far apart, close together, form a ray, a line, or a line segment. If you want to show that three points are collinear, choose two line segments, for example, PQ and QR. We, then, need to establish that they have a common direction (that is, equal gradients) and a common point (for example, Q). If both of these statements are true then the points can be considered as collinear.
What is the Difference Between Collinearity and Coplanarity?
In geometry, three or more points are considered to be collinear if they all lie on a single straight line. This property of the points is called collinearity. On the other hand, three or more points are considered to be coplanar if they all lie in the same plane. This property of the points is called coplanarity.
Can Collinear Points be Coplanar As Well?
Yes, we can consider collinear points to be coplanar as well. If the points are collinear then we can choose any one of an infinite number of planes that contains the line on which all these points lie. Hence, collinear points are coplanar, but coplanar points may not necessarily be collinear.