Worksheet on collinear points is much useful to the students who would like to practice problems on collinearity.

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

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

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

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

**Problem 1 :**

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 "Condition of collinearity of three points"

**Problem 2 :**

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 "Condition of collinearity of three points"

**Problem 3 :**

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.

**Problem 4 :**

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 "Condition of collinearity of three points"**

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