What are Collinear Points in Geometry – Definition, Properties & Examples

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A point is one of the basic and fundamental shapes in mathematics. A point can be of different types depending on its position relative to other points or figures. One such category of points is collinear points.

Let’s understand what are collinear points in geometry and their properties with examples.

What are Collinear Points in Geometry?

The term collinear is derived from two Latin words ‘col’ and ‘linear’. ‘Col’ means together and ‘Linear’ means line. Therefore, collinear points are the points that lie on a single line. 

You can find many real-life examples of collinearity such as a group of students standing in a straight line, books kept in a row in a library, items arranged in a row in a grocery store, etc.

In geometry, two or more points are said to be collinear, if they lie on the same line. Hence the collinear points are the set of points that lie on a single straight line.

Examples

In the figure below, the three points A, B, and C are called collinear points as they all lie on a single line. In other words, we can say that we can draw a line passing through all three points.

Note: 

• Two points are always collinear points.
• A triangle cannot be formed using three collinear points.

Let’s consider one more figure.

In the above figure, the points P, Q, and R are non-collinear points, as they all three are not lying on a single line. In other words, we can say that we cannot draw a line passing through all three points.

Note: Three non-colinear points always form a triangle.

How to Check Three Points Are Collinear?

As discussed above, three points are called collinear points, if they all lie on the same line. In order to check, whether the three given points are collinear, any of the following three methods can be used.

• Distance Formula Method
• Slope Formula Method
• Area of Triangle Method

Distance Formula Method

In this method, we find the distance between the first and the second point, and then the distance between the second and the third point. After this, we check if the sum of these two distances is equal to the distance between the first and the third point. This will only be possible if the three points are collinear points. 

To calculate the distance between two points whose coordinates are known to us, we use the distance formula.

The distance between two points $\text{A} \left(x_1, y_1 \right)$ and $\text{B} \left(x_2, y_2 \right)$ is $\sqrt{\left(x_2 – x_1 \right)^2 + \left(y_2 – y_1 \right)^2}$

So, if we have three collinear points in the order A, B, and C, then these points will be collinear if AB + BC = CA.

Let’s consider a few examples to understand the method.

Examples

Ex 1: Check whether the points $\text{P}(−3,−1)$, $\text{Q}(−1,0)$, and $\text{R}(1,1)$ are collinear.

Distance between the points P and Q = |PQ| = $\sqrt{\left( -1 – (-3)\right)^2 + \left(0 – (-1) \right)^2}$

$= \sqrt{\left( -1 + 3\right)^2 + \left(0 + 1 \right)^2}$

$= \sqrt{2^2 + 1^2}$

$= \sqrt{4 + 1}$

$= \sqrt{5} \text{ units}$

Distance between the points Q and R = |QR| = $\sqrt{\left(1 – (-1)\right)^2 + \left(1 – 0 \right)^2}$

$= \sqrt{\left(1 + 1\right)^2 + \left(1 – 0 \right)^2}$

$= \sqrt{2^2 + 1^2}$

$= \sqrt{4 + 1}$

$= \sqrt{5} \text{ units}$

Distance between the points P and R = |PR| = $\sqrt{\left(1 – (-3)\right)^2 + \left(1 – \left(-1 \right) \right)^2}$

$= \sqrt{\left(1 + 3\right)^2 + \left(1 + 1 \right)^2}$

$= \sqrt{\left(4\right)^2 + \left(2 \right)^2}$

$= \sqrt{16 + 4}$

$= \sqrt{20}$

$= 2\sqrt{5} \text{ units}$

Note that $\sqrt{5} + \sqrt{5} = 2\sqrt{5}$, therefore, PQ + QR = PR, which means that the three points P, Q, and R are collinear points.

Ex 2: Check whether the points $\text{A}(1,0)$, $\text{B}(0,0)$, and $\text{C}(0,1)$ are collinear.

Distance between the points A and B = |AB| = $\sqrt{\left( 0 – 1\right)^2 + \left(0 – 0 \right)^2}$

$ = \sqrt{\left(- 1\right)^2 + 0^2}$

$ = \sqrt{1 + 0}$

$ = \sqrt{1}$

$ = 1 \text{ units}$

Distance between the points B and C = |BC| = $\sqrt{\left(0 – 0\right)^2 + \left(1 – 0 \right)^2}$

$= \sqrt{0^2 + 1^2}$

$= \sqrt{0 + 1}$

$= \sqrt{1}$

$ = 1 \text{ units}$

Distance between the points A and C = |AC| = $\sqrt{\left(0 – 1\right)^2 + \left(1 – 0 \right)^2}$

$=\sqrt{\left(- 1\right)^2 + 1^2}$

$=\sqrt{1 + 1}$

$=\sqrt{2} \text{ units}$

Since the length AC is greater than the other two, therefore, we will add the other two distances, i.e., AB and BC.

AB + BC = $1 + 1 = 2 \text{ units}$

Since, S\sqrt{2} \ne 2$ and hence, AB + BC $\ne$ AC, therefore, the points A, B, and C are not collinear points. Points A, B, and C are non-collinear points.

Slope Formula Method

Three or more points are said to be collinear if the slope of any two pairs of points is the same. The slope of the line basically measures the steepness of the line.

Suppose, A, B, and C are the three points, with which we can form three sets of pairs, such that, AB, BC, and CA are three pairs of points. Then, as per the slope formula,

If the slope of AB = Slope of BC = Slope of CA, then the points A, B, and C are collinear.

Slope of the line segment joining two points $\left(x_1, y_1 \right)$ and $\left(x_2, y_2 \right)$ is given by the formula $m = \frac{y_2 – y_1}{x_2 – x_1}$.

Let’s consider a few examples to understand the method.

Examples

Ex 1: Check whether the points $\text{P}(−3,−1)$, $\text{Q}(−1,0)$, and $\text{R}(1,1)$ are collinear.

Slope of line PQ = $\frac{0 – (-1)}{-1 – (-3)} = \frac{0 + 1}{-1 + 3} = \frac{1}{2}$

Slope of line QR = $\frac{1 – 0}{1 – (-1)} = \frac{1}{1 + 1} = \frac{1}{2}$

Since the slopes of the lines, PQ and QR are equal, therefore, the points P, Q, and R are collinear points.

Ex 2: Check whether the points $\text{A}(1,0)$, $\text{B}(0,0)$, and $\text{C}(0,1)$ are collinear.

Slope of line AB = $\frac{0 – 0}{0 – 1} = \frac{0}{-1} = 0$

Slope of line BC = $\frac{1 – 0}{0 – 0} = \frac{1}{0} = \text{ undefined}$

Since the slopes of the lines, AB and BC are not equal, therefore, the points A, B, and C are non-collinear points.

Area of Triangle Method

To check whether the three points are collinear, we find the area of a triangle using the three points. If 

• area of a triangle is $0$, then the three points are collinear
• area of a triangle is non-zero, then the three points are non-collinear

To find the area of a triangle with vertices $\text{A}\left(x_1, y_1 \right)$, $\text{B}\left(x_2, y_2 \right)$, and $\text{C}\left(x_3, y_3 \right)$, we use the formula $\frac{1}{2} \left(x_1 \left(y_2 – y_3\right) + x_2 \left(y_3 – y_1\right) + x_3 \left(y_1 – y_2\right) \right)$.

Note: When three points are collinear, a triangle cannot be formed, and hence, the area of the triangle, in this case, is $0$.

Let’s consider a few examples to understand the method.

Examples

Ex 1: Check whether the points $\text{P}(−3,−1)$, $\text{Q}(−1,0)$, and $\text{R}(1,1)$ are collinear.

The area of a triangle formed using the points P, Q, and R is $\frac{1}{2} \left(-3 \left(0 – 1\right) + (-1) \left(1 – (-1)\right) + 1 \left(-1 – 0\right) \right)$

$= \frac{1}{2} \left(-3 \left(-1\right) + (-1) \left(1 + 1\right) + 1 \left(-1 – 0\right) \right)$

$= \frac{1}{2} \left(-3 \times \left(-1\right) + (-1) \times 2 + 1 \times \left(-1\right) \right)$

$= \frac{1}{2} \left(3 – 2 – 1 \right) = 0 \text{ sq units}$.

Since, the area of a triangle formed by the points $\text{P}(−3,−1)$, $\text{Q}(−1,0)$, and $\text{R}(1,1)$  is $0 \text{sq units}$, therefore, the point P, Q, and R are collinear.

Ex 2: Check whether the points $\text{A}(1,0)$, $\text{B}(0,0)$, and $\text{C}(0,1)$ are collinear.

The area of a triangle formed using the points P, Q, and R is $\frac{1}{2} \left(1 \left(0 – 1\right) + 0 \left(1 – 0\right) + 0 \left(0 – 0\right) \right)$

$= \frac{1}{2} \left(1 \times \left(- 1\right) + 0 \times 1 + 0 \times 0 \right)$

$= \frac{1}{2} \left(-1 + 0 \times 1 + 0 \times 0 \right)$

$= \frac{1}{2} \left(-1 + 0 + 0 \right)$

$= \frac{1}{2} \left(-1 \right)$

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

As the area of a triangle cannot be negative, therefore, area of the triangle is $1 \text{ sq units}$.

Since, the area of a triangle formed by the points $\text{A}(1,0)$, $\text{B}(0,0)$, and $\text{C}(0,1)$  is non-zero, therefore, the point A, B, and C are non-collinear.

Practice Problems

1. What are collinear points?
2. What are non-collinear points?
3. What are the three methods to check whether the three points are collinear?
4. Check whether the points P(1, 2), Q(2, 3), and R(3, 4) are collinear points or not using
   • distance formula method
   • slope formula method
   • area of the triangle method
5. Check whether the points P(-3, -1), Q(-1, 2), and R(1, 1) are collinear points or not using
   • distance formula method
   • slope formula method
   • area of the triangle method

FAQs

Conclusion

Collinear points are a set of three or more points that exist on the same straight line. The points that are not collinear are called non-collinear points. The three methods to check the collinearity of three points are using the distance formula, the slope formula, and the area of triangle formula.

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