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If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

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Given: A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)

To prove: A, B and C are collinear

To find: the ratio in which C divides AB

Formula used: 

Section Formula: 

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

\(\Big(\frac{nx+ma}{m+n},\frac{ny+mb}{m+n},\frac{nz+mc}{m+n}\Big)\) 

Let C divides AB in ratio k: 1 

Three points are collinear if the value of k is the same for x, y and z coordinates 

Therefore, m = k and n = 1 

A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)

Coordinates of C using section formula:

On comparing:

\(\frac{9k+3}{k+1}\) = 5

⇒ 9k + 3 = 5(k + 1)

⇒ 9k + 3 = 5k + 5 

⇒ 9k – 5k = 5 – 3

⇒ 4k = 2

⇒ k = \(\frac{2}{4}\) 

⇒ k = \(\frac{1}{2}\) 

\(\frac{8k+2}{k+1}\) = 4

⇒ 8k + 2 = 4(k + 1) 

⇒ 8k + 2 = 4k + 4 

⇒ 8k – 4k = 4 – 2 

⇒ 4k = 2

⇒ k = \(\frac{2}{4}\) 

⇒ k = \(\frac{1}{2}\) 

\(\frac{-10k-4}{k+1}\) = -6

⇒ -10k – 4 = -6(k + 1) 

⇒ -10k – 4 = -6k – 6 

⇒ -10k + 6k = 4 – 6 

⇒ -4k = -2

⇒ k = \(\frac{-4}{-2}\) 

⇒ k = \(\frac{1}{2}\) 

The value of k is the same in all three times 

Hence, A, B and C are collinear

As,  k = \(\frac{1}{2}\) 

C divides AB externally in ratio 1:2

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