Question

PROVE THAT THE POINT (3,-2) , (-5,4) AND (-1,1) are collinear.
PROVIDE FULL EXPLANATION PLEASE. ​

Answer 1

→ Given ←

3 points (3, -2) , (-5, 4) and (-1, 1).

Let -

• Point A be (3, -2).
• Point B be (-5, 4).
• Point C be (-1, 1).

→ To Prove ←

ABC is co-linear.

→ Proof ←

We know,

Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

We have :-

• x₁ = 3 & y₁ = -2
• x₂ = -5 & y₂ = 5
• x₃ = -1 & y₃ = 1

Substituting the values :-

A = 1/2 { 3(5 - 1) + (-5)[1 - (-2)] + (-1)(-2 - 1) }

→ A = 1/2 [ (3 × 4) + (-5 × 3) + (-1 × -3) ]

→ A = 1/2 [ 12 - 15 + 3 ]

→ A = 1/2 (15 - 15)

→ A = 1/2 (0)

→ A = 0/2

→ A = 0 sq. units

Hence the given points are co-linear.

Proved !

Answer 2

Answer:

$\huge\fcolorbox{black}{aqua}{Solution:-}$

→ Given ←

3 points (3, -2) , (-5, 4) and (-1, 1).

Let -

Point A be (3, -2).

Point B be (-5, 4).

Point C be (-1, 1).

→ To Prove ←

ABC is co-linear.

→ Proof ←

We know,

Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

We have :-

x₁ = 3 & y₁ = -2

x₂ = -5 & y₂ = 5

x₃ = -1 & y₃ = 1

Substituting the values :-

A = 1/2 { 3(5 - 1) + (-5)[1 - (-2)] + (-1)(-2 - 1) }

→ A = 1/2 [ (3 × 4) + (-5 × 3) + (-1 × -3) ]

→ A = 1/2 [ 12 - 15 + 3 ]

→ A = 1/2 (15 - 15)

→ A = 1/2 (0)

→ A = 0/2

→ A = 0 sq. units

$</p><p>\fcolorbox{black}{aqua}{Hence the given points are co-linear}$

Proved !