Question

Find x' if the pts are colliner
(x-1), (2, 1) and (4,5)​

Answer 1

Answer:

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Answer 2

• For the Points (Vertices of Triangle) to be Collinear, The Area of Triangle should be 0

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

We have :

• x₁ = x , x₂ = 2 , x₃ = 4

• y₁ = -1 , y₂ = 1 , y₃ = 5

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

⇒ 0 =  1/2 [x (1 - 5 ) + 2 (5 - (-1) ) + 4((-1) - 1)]

⇒ 0 =  1/2 [x (-4) + 2 (5 + 1) + 4(-1 - 1)]

⇒ 0 =  1/2 [-4x + 2 (6) + 4(-2)]

⇒ 0 =  1/2 [-4x + 12 + (-8)]

⇒ 0 =  1/2 [-4x + 12 - 8]

⇒ 0 =  1/2 [-4x + 4]

⇒ 0 =  (-4x + 4)/2

Multiplying both sides by 2

⇒ 0 × 2 =  (-4x + 4)/2 × 2

⇒ 0  =  (-4x + 4)

⇒ 0  =  -4x + 4

⇒ 0 - 4  =  -4x + 4 - 4

Subtracting 4 from both sides

⇒ - 4  =  -4x

Dividing -4 from both sides

⇒ - 4/-4  =  -4x/-4

⇒ 1 =  x

Switch Sides

⇒ x = 1