Given: A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)
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(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)
Coordinates of C using section formula:
⇒ (-4,1,-10) = \(\Big(\frac{k(-1)+1(2)}{k+1},\frac{k(2)+1(3)}{k+1},\frac{k(-3)+1(4)}{k+1}\Big)\)
⇒ (-4,1,-10) = \(\Big(\frac{-k+2}{k+1},\frac{2k+3}{k+1}, \frac{-3k+4}{k+1}\Big)\)
On comparing:
\(\frac{-k+2}{k+1}\) = - 4
⇒ -k + 2 = -4(k + 1)
⇒ -k + 2 = -4k – 4
⇒ 4k – k = - 2 – 4
⇒ 3k = - 6
⇒ k = \(\frac{-6}{3}\)
⇒ k = - 2
\(\frac{2k+3}{k+1}\) = 1
⇒ 2k + 3 = k + 1
⇒ 2k – k = 1 – 3
⇒ k = – 2
\(\frac{-3k+4}{k+1}\) = -10
⇒ -3k + 4 = -10(k + 1)
⇒ -3k + 4 = -10k – 10
⇒ -3k + 10k = -10 – 4
⇒ 7k = -14
⇒ k = \(\frac{-14}{7}\)
⇒ k = - 2
The value of k is the same in all three times
Hence, A, B and C are collinear
As k = -2
C divides AB externally in ratio 2:1
