Here we will learn about condition of collinearity of three 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**

**Form Condition of Collinearity of Three Points to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.