Condition of Collinearity of Three Points

Here we will learn about condition of collinearity of three points.

How to find the condition of collinearity of three given points?

First Method:

Let us assume that the three non-coincident points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are collinear. Then, one of these three points will divide the line segment joining the other two internally in a definite ratio. Suppose, the point B divides the line segment  AC  internally in the ratio λ : 1. 

Hence, we have, 

(λx₃ + 1 ∙ x₁)/(λ + 1) = x₂ …..(1) 

and (λy₃ + 1 ∙ y₁)/(λ+1) = y₂ ..…(2) 

From (1) we get, 

λx₂ + x₂ = λx₃ + x₁

or, λ (x₂ - x₃) = x₁ - x₂

or, λ = (x₁ - x₂)/(x₂ - x₃)

Similarly, from (2) we get, λ = (y₁ - y₂)/(y₂ - y₃)

Therefore, (x₁ - x₂)/(x₂ - x₃) = (y₁ -y₂)/(y₂ - y₃)

or, (x₁ - x ₂)(y₂ - y₃) = (y₁ - y₂) (x₂ - x₃ )

or, x₁ (y₂ - y₃) + x₂ y₃ - y₁) + x₃ (y₁ - y₂) = 0

which is the required condition of collinearity of-the three given points.

Second Method: 

Let A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃)be three non-coincident points and they are collinear. Since area of a triangle = ½ ∙ base × altitude, hence it is evident that the altitude of the triangle ABC is zero, when the points A, B, and C are collinear. Thus, the area of the triangle is zero if the points A, B and Care collinear. Therefore, the required condition of collinearity is

1/2 [x₁ (y₂ - y₃) + x₂(y₃ - y₁) + x₃ (y₁ - y₂)] = 0

or, x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂) = 0.

Examples on Condition of Collinearity of Three Points: 

1. Show that the points (0, -2) , (2, 4) and (-1, -5) are collinear. 

Solution:

The area of the triangle formed by joining the given points

= 1/2 [(0 - 10 + 2) - (-4 -4 + 0)] = 1/2 (-8 + 8) = 0.

Since the area of the triangle formed by joining the given points is zero, hence the given points are collinear. Proved

2. Show that the straight line joining the points (4, -3) and (-8, 6) passes through the origin.

Solution:

The area of the triangle formed by joining the points (4, -3), (-8, 6) and (0, 0) is 1/2 [24 - 24] = 0.

Since the area of the triangle formed by joining the points (4, -3), (-8, 6) and (0, 0) is zero, hence the three points are collinear : therefore, the straight line joining the points (4, -3) and (-8, 6)passes through the origin.

3. Find the condition that the points (a, b), (b, a) and (a², – b²) are in a straight line.

Solution:

Since the three given points are in a straight line, hence the area of the triangle formed by the points must be zero.

Therefore, 1/2 | (a² - b³ + a²b) – (b² + a³ - ab²) | = 0

or, a² - b³ + a²b – b² – a³ + ab² = 0

or, a² – b² – (a³ + b³) + ab (a + b) = 0

or, (a + b) [a - b - (a² - ab + b²) + ab] = 0

or, (a + b) [(a - b)- (a² - ab + b² - ab)] = 0

or, (a + b) [(a - b) - (a - b)²] = 0

or, (a + b) (a - b) (1 - a + b) = 0

Therefore, either a + b = 0 or, a – b = 0 or, 1 - a + b = 0.

● Co-ordinate Geometry 

• What is Co-ordinate Geometry?
  
• Rectangular Cartesian Co-ordinates
  
• Polar Co-ordinates
  
• Relation between Cartesian and Polar Co-Ordinates
  
• Distance between Two given Points
  
• Distance between Two Points in Polar Co-ordinates
  
• Division of Line Segment: Internal & External
  
• Area of the Triangle Formed by Three co-ordinate Points
  
• Condition of Collinearity of Three Points
  
• Medians of a Triangle are Concurrent
  
• Apollonius' Theorem
  
• Quadrilateral form a Parallelogram 
  
• Problems on Distance Between Two Points 
  
• Area of a Triangle Given 3 Points
  
• Worksheet on Quadrants
  
• Worksheet on Rectangular – Polar Conversion
  
• Worksheet on Line-Segment Joining the Points
  
• Worksheet on Distance Between Two Points
  
• Worksheet on Distance Between the Polar Co-ordinates
  
• Worksheet on Finding Mid-Point
  
• Worksheet on Division of Line-Segment
  
• Worksheet on Centroid of a Triangle
  
• Worksheet on Area of Co-ordinate Triangle
  
• Worksheet on Collinear Triangle
  
• Worksheet on Area of Polygon
  
• Worksheet on Cartesian Triangle

11 and 12 Grade Math

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