Question

Show that A (4,1), B (5,-2) and c (6,-5) are collinear. [Note: slope of AB = slope of BC= slope of AC​​

Answer 1

Step-by-step explanation:

Given :-

The points A (4,1), B (5,-2) and C (6,-5)

Required To Prove:-

A , B and C are collinear points .

Solution :-

Given points are A (4,1), B (5,-2) and C (6,-5)

To show that A,B and C are collinear then we have to show that slope of AB = slope of BC = slope of AC

Finding Slope of AB :-

Let (x₁, y₁) = (4,1) => x₁ = 4 and y₁ = 1

Let ( x₂,y₂) = (5,-2) => x₂ = 5 and y₂ = -2

We know that

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

(y₂-y₁)/(x₂-x₁)

Slope of the line segment AB

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

= -3/1

= -3

Therefore, Slope of AB = -3 ----(1)

Finding Slope of BC :-

Let (x₁, y₁) = (5,-2) => x₁ = 5 and y₁ = -2

Let ( x₂,y₂) = (6,-5) => x₂ = 6 and y₂ = -5

We know that

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

(y₂-y₁)/(x₂-x₁)

Slope of the line segment BC

= [-5-(-2)]/(6-5)

= (-5+2)/1

= -3/1

= -3

Therefore, Slope of BC = -3 ----(2)

Finding Slope of AC :-

Let(x₁, y₁) = (4,1) => x₁ = 4 and y₁ = 1

Let ( x₂,y₂) = (6,-5) => x₂ = 6 and y₂ = -5

We know that

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

(y₂-y₁)/(x₂-x₁)

Slope of the line segment AC

= (-5-1)/(6-4)

= -6/2

= -3

Therefore, Slope of AC = -3 ------(3)

From (1) , (2) & (3)

slope of AB = slope of BC = slope of AC

Therefore, A,B and C are collinear points.

Hence, Proved.

Used formulae:-

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

(y₂-y₁)/(x₂-x₁)

Used Concept :-

If three points A, B and C are collinear then slope of AB = slope of BC = slope of AC

Points to know:-

→ The points lie on the same line are called collinear points.

→ If A, B and C are collinear then AB+BC = AC.

→ The area of a triangle of collinear points is zero. Because no triangle is formed by the collinear points.

Answer 2

$\huge\mathfrak\pink{Answer}$

Slope of AB= 12−8/5-4

= 4/1=4

Slope of BC= 28−12/9-5

= 16/4 =4

Hence AB∣∣BC, but B is a point of intersect.

Hence A, B, C are collinear.

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