Class 10 Maths- Collinearity of 3 points

Collinearity of 3 points

Three points (x₁, y₁), (x₂, y₂) and (x₃, y₃) are collinear if

x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) = 0

(x₁y₂ + x₂y₃ + x₃y₁) – (x₁y₃ + x₃y₂ + x₂y₁) = 0

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

Let A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃)be three non-coincident points and they are also 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.

Let us understand with examples:

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. 

Example: 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.