**Conditions for collinearity of three points :**

The following are the conditions for collinearity of three points.

Let A, B and C be the three points.

If we want A, B and C be collinear, the following conditions have to be met.

(i) Slope of AB = Slope of BC

(ii) There must be a common point between AB and BC.

(In AB and BC, the common point is B)

If the above two conditions are met, then the three points A, B and C are collinear.

Let A, B and C be the three points.

We have to find the three lengths AB, BC and AC among the given three points A, B and C.

The three points A, B and C are collinear, if the sum of the lengths of any two line segments among AB, BC and AC is equal to the length of the remaining line segment.

That is,

AB + BC = AC

(or)

AB + AC = BC

(or)

AC + BC = AB

Let A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) be the three points.

If the three points, A, B and C are collinear, they will lie on the same and they cannot form a triangle.

Hence, the area of triangle ABC = 0

1/2 x { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) } = 0

(or)

**x₁y₂ + x₂y₃ + x₃y₁ = x₂y₁ + x₃y₂ + x₁y₃**

Let A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) be the three points.

Let us find the equation through any two of the given three points. If the third point satisfies the equation, then the three points A, B and C will be collinear.

**Problem 1 :**

Using the concept of distance between two points, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

**Solution :**

We know the distance between the two points (x₁, y₁) and (x₂, y₂) is

d = √ (x₂ - x₁) ² + (y₂ - y₁) ²

Let us find the lengths AB, BC and AC using the above distance formula.

AB = √ [(4 - 5)² + (-1 + 2)²]

AB = √ [(-1)² + (1)²]

AB = √ [1 + 1]

AB = √2

BC = √ [(1 - 4)² + (2 + 1)²]

BC = √ [(-3)² + (3)²]

BC = √ [9 + 9]

BC = √[2X9]

BC = 3√2

AC = √ [(1 - 5)² + (2 + 2)²]

AC = √ [(-4)² + (4)²]

AC = √ [16 + 16]

AC = √[2x16]

AC = 4√2

Therefore, AB + BC = √2 + 3√2 = 4√2 = AC

Thus, **AB + BC = AC**

**Hence, the given three points A, B and C are collinear.**

Let us look at the next problem on "Conditions for collinearity of three points"

**Problem 2 :**

Using the concept of slope, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

**Solution :**

Slope of the line joining (x₁, y₁) and (x₂, y₂) is,

m = (y₂ - y₁) / (x₂ - x₁)

Using the above formula,

Slope of the line AB joining the points A (5, - 2) and B (4- 1) is

= (-1 + 2) / (4 - 5)

= - 1

Slope of the line BC joining the points B (4- 1) and C (1, 2) is

= (2 + 1) / (1 - 4)

= - 1

Thus, **slope of AB = slope of BC**.

And also, **B is the common point**.

**Hence, the points A , B and C are collinear.**

Let us look at the next problem on "Conditions for collinearity of three points"

**Problem 3 :**

Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

**Solution :**

Using the concept of area of triangle, if the three points A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are collinear, then

x₁y₂ + x₂y₃ + x₃y₁ = x₂y₁ + x₃y₂ + x₁y₃

Comparing the given points to A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃), we get

(x₁, y₁) = (5, -2)

(x₂, y₂) = (4, -1)

(x₃, y₃) = (1, 2)

x₁y₂ + x₂y₃ + x₃y₁ = 5x(-1) + 4x2 + 1x(-2)

x₁y₂ + x₂y₃ + x₃y₁ = -5 + 8 -2

**x₁y₂ + x₂y₃ + x₃y₁ = 1 --------(1)**

x₂y₁ + x₃y₂ + x₁y₃ = 4x(-2) + 1x(-1) + 5x(2)

x₂y₁ + x₃y₂ + x₁y₃ = -8 - 1 + 10

**x₂y₁ + x₃y₂ + x₁y₃ = 1 --------(2)**

From (1) and (2), we get

**x₁y₂ + x₂y₃ + x₃y₁ = x₂y₁ + x₃y₂ + x₁y₃**

**Hence, the three points A, B and C are collinear.**

Let us look at the next problem on "Conditions for collinearity of three points"

**Problem 4 :**

Using the concept of equation of line, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

**Solution :**

Equation of the straight line in two-points form is

(y - y₁) / (y₂ - y₁) = (x - x₁) / (x₂ - x₁)

Using the above formula, let us get equation of the line through the points A and B.

Plugging (x₁ , y₁) = (5, -2) and (x₂, y₂) = (4, -1), we get

(y +2) / (-1 + 2) = (x - 5) / (4 - 5)

(y + 2) / 1 = (x - 5) / (-1) ----------> y +2 = -x + 5

y +2 = -x + 5 ----------> x + y - 3 = 0

Now, we can plug the third point C(1, 2) in the above equation.

That is, plug x = 1 and y = 2

x + y - 3 = 0 ----------> 1 + 2 - 3 = 0 -------> 0 = 0

Therefore, the third point C(1, 2) satisfies the equation.

Hence, the given points three points A, B and C are collinear.

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