Collinear Point Formula, Properties and Questions

Collinear points are three or more points that lie on the same line, whether they are close together, far apart, or form a ray, line segment, or line. Let us understand this with the help of an example of students standing in a line for their morning assembly. 

Here, P, Q, and R are collinear points as they lie on the same straight line.

3 or more points that do not lie on the same line, and do not connect to form a straight line, are called non-collinear points. For example, points on spirals, points on the globe, etc. With non-collinear points, it is possible to form various shapes like a triangle, a quadrilateral, a parallelogram, etc. JEE Main entrance exam and IIT JAM examination both ask questions related to the non-collinearity of points.

 Here, X, Y, and Z do not lie on the same line and are thus non-collinear points.

The IISER entrance exam requires students to know about the properties of the collinear points. Based on these properties, questions are formulated for the entrance exam. The following points explain the properties of collinear points:

• Every collinear point lies on a straight line.
• The slope between any two pairs of collinear points is the same. This is proven by the slope formula.
• The area of a triangle for collinear points will be zero.
• The distance formula is used for calculating the distance between collinear points. The distance formula is $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

• Collinear points are represented as scalar multiples of the direction vector.
• If there is a vector between two points, any other collinear point is represented as a scaled version of this vector.
• In linear algebra, collinear points are linearly dependent. One of the points can be expressed as a linear combination of others.

There are different formulae that are used for determining collinear points. Questions based on these formula are often asked in NEET exam and CUET exam. Let us learn more about them. 

1. Area of Triangle Formula

This method is based on the fact that collinear points cannot create a triangle. If by including three points, the area comes out to be zero, then those points will be collinear. Let us say that there are three points $A(x_1,y_1),B(x_2,y_2),\text{and}C(x_3,y_3)$ . Here, the area of the triangle will be:

$A=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|=0$

This proves that these three points are collinear since the area of the triangle comes out to be zero.

2. Slope Formula

Through this method, collinearity is confirmed if the slopes of all three points is zero. In simple words, if the slopes of all the points (which should be 3 or more) is zero, then all those points are collinear. Let us understand this with an example. Say we have three points X, Y and Z. First, the slope formula is applied on X and Y. For x and y, the slope formula between $(x_1,y_1)\text{and}(x_2,y_2)\; is\; given\; by:$

$m=\frac{y_2-y_1}{x_2-x_1}$

Say there are three points A(1,2), B(2,4) and C(3,6). We will first calculate the slope between A and B. 

$m_{AB}=\frac{4-2}{2-1}=\frac{2}{1}=2$

After that, we will calculate the slope between B and C.

$m_{BC}=\frac{6-4}{3-2}=\frac{2}{1}=2$

As we can see, for all three points, the slope came out to be equal, i.e. 2. Therefore, A, B and C are all collinear.

Let us consider some questions that are often asked in school and entrance examinations:

1. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.

Solution. The points are collinear if the sum of any two line segments is equal to the third.

Let A (1, 5), B (2, 3), and C (-2, -11).

AB= √{(2-1)2 + (3-5)2} = √5

BC= √{(-2-2)2 + (-11-3)2}= √212

CA= √{(-2-1)2 + (-11-5)2} = √265

AB + BC ≠ CA.

Hence, the given points are not collinear.

2. Find the coordinates of the point that divides the join of (V1, 7) and (4, V3) in the ratio. 

Solution. Let the required point be P (x, y).

Using section formula, 

x= (2*4+3*(-1))/(2+3) = 1,

y= (2*-3+3*7)/(2+3)= 3.

The point is (1,3).

3. Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).

Solution.  Let k:1 be the ratio in which the line segment is divided.

Thus, -1= (6k-3)/(k+1),

-k-1=6k-3,

k=2/7.

The required ratio is 2:7.