Collinear points with Examples and FAQs

What are Collinear points?

Three or more points can be identified to be collinear if a straight line passes through all the points. This line is unique. The condition for three or more points to be collinear is known as collinearity.

Set of three collinear points A, B and C

The above graph represents three points A, B and C being plotted with respect to their x and y coordinates. A line can be drawn such that all the three points lie on the line, this line is unique. Hence we can say that points A, B and C are collinear.

The coordinates of the collinear points will satisfy the equation of the single straight line passing through them.   

Non-Collinear Points

When it is not possible to connect three or more points using a single straight line, then such points are termed to be non-collinear.

Set of four non-collinear points

The above graph represents four points A, B, C and D being plotted with respect to their x and y coordinates. No such line can be drawn that connects all the four points as shown on the coordinate plane. Hence we can say that points A, B, C and D are non-collinear.

A straight line passing through the points can be easily deduced visually. But there are some formulas to verify if the given points are collinear or non-collinear. Let’s discuss these in detail.

Formula to check collinearity of points:

1. Distance Formula:

Let X, Y and Z be three points whose collinearity needs to be verified.

If the three given points are collinear then a straight line must pass through them and hence the sum of the distance between X and Y i.e., XY and the distance between Y and Z i.e., YZ will be equal to the distance between X and Z i.e., XZ.

⇒ XY + YZ = XZ

We know, that the distance formula for two points having coordinates (x, y) and (x’, y’) is:

\text { Distance }=\sqrt{\left(x^{\prime}-x\right)^{2}+\left(y^{\prime}-y\right)^{2}}

Hence using this formula, the distance between XY, YZ and XZ can be easily calculated and then equated in the equation: XY + YZ = XZ, to check collinearity.  

2. Section Formula:

Let P, Q and R be three points whose collinearity needs to be verified.

If the three given points are collinear then a straight line will pass through all three of them. 

Let us consider the line segment joining P and R will be divided by the point Q internally in the ratio m : n.

The section formula is:

If A\left(x_{1}, y_{1}\right) \text { and } B\left(x_{2}, y_{2}\right) are two points and a point O(x,y) divides the line segment AB is the ratio m:n then we have

(x, y)=\left(\frac{m x_{1}+n x_{2}}{m+n}, \frac{m y_{1}+n y_{2}}{m+n}\right)

To prove the collinearity of three points using the section formula we will verify if the point Q divides the line joining PR in a ratio m:n. 

Area of Triangle Formula:

If three points are collinear then they represent points on a straight line. Hence, for any three points if the area of the triangle is calculated to be zero, then the points will be collinear.

We know the formula for the area of a triangle, i.e.:

If A(x_1,y_1),B(x_2,y_2)\text{ and }C(x_3,y_3) are three points that form a triangle then area is: 

Area of a triangle =\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]

If the value of the area of a triangle is zero, then we can say that the three points are collinear.

Examples

Example 1: From the figure given below mention the points that exhibit collinearity.

Solution:

The points that are collinear are 

A and B.

A, C, D and E.

B, F and H.

E, F and G.

Example 2: Check if the points (1, 5), (3, 3) and (6, 0) are collinear.

Solution:

Let us name the points (1, 5), (3, 3) and (6, 0) as X, Y and Z respectively.

Now to prove the collinearity we must have

XY + YZ = XZ

Where XY is the distance between X and Y and similarly for YZ and XZ.

Using distance formula for two points

\text { Distance }=\sqrt{\left(x^{\prime}-x\right)^{2}+\left(y^{\prime}-y\right)^{2}}

We have,

X Y=\sqrt{(3-1)^{2}+(3-5)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=2 \sqrt{2}
Y Z=\sqrt{(6-3)^{2}+(0-3)^{2}}=\sqrt{\left(3^{2}\right)+(-3)^{2}}=\sqrt{9+9}=3 \sqrt{2}
X Z=\sqrt{(6-1)^{2}+(0-5)^{2}}=\sqrt{(5)^{2}+(-5)^{2}}=\sqrt{25+25}=5 \sqrt{2}
\therefore X Y+Y Z=2 \sqrt{2}+3 \sqrt{2}=5 \sqrt{2}=X Z

Hence the given points (1, 5), (3, 3) and (6, 0) are collinear.

Example 3: Verify if the points A(3, -3), B(6, 0) and C(9, 3) are collinear using section formula. 

Solution:

if A\left(x_{1}, y_{1}\right) \text { and } B\left(x_{2}, y_{2}\right) are two points and a point O(x,y) divides the line segment AB is the ration of m:n  the we have 

(x, y)=\left(\frac{m x_{1}+n x_{2}}{m+n}, \frac{m y_{1}+n y_{2}}{m+n}\right)

Let the ratio in which B divides the line segment AC be p : 1.

Now by the section formula, we have the coordinates of B is:

\left(\frac{3 p+9}{p+1}, \frac{-3 p+3}{p+1}\right)

The given coordinates of B is (6, 0)

Now comparing the value for x and y coordinates and solving for p:

\frac{3 p+9}{p+1}=6 \text { and } \frac{-3 p+3}{p+1}=0

\frac{3 p+9}{p+1}=6 \Rightarrow 3 p+9=6 p+6

\Rightarrow 3 p=3
\Rightarrow p=1

\text { and } \frac{-3 p+3}{p+1}=0 \Rightarrow-3 p+3=0

\Rightarrow-3 p=-3
\Rightarrow p=1

The value of p is the same after solving. Hence the points A(3, -3), B(6,0) and C(9,3) are collinear.

Example 4: Check if the points (1, 4), (3, 2) and (6, -1) are collinear using the area of triangle formula.

Solution:

If A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) \text { and } C\left(x_{3}, y_{3}\right) are three points that  form a triangle then area is : 

Area of a triangle =\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]

To prove if the points are collinear the area must be equal to zero.

The area of the triangle formed by points (1, 4), (3, 2) and (6, -1) is:

Area of a triangle 

= \frac{1}{2}[1(2-(-1))+3(-1-4)+6(4-2)]

= \frac{1}{2}[1(2+1)+3(-5)+6(2)]

= \frac{1}{2}[3-15+12]

= 0

Hence the given points (1, 4), (3, 2) and (6, -1) are collinear.