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
