Question

Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and are collinear.

Answer 1

The given points are A=( 2,−3,4 ) , B=( −1,2,1 ) and C=( 0, 1 3 ,2 ) .

Let the point Q=( x,y,z ) divides AB in the ratio k:1 .

According to section formula,

Q( x,y,z )=( m x 2 +n x 1 m+n , m y 2 +n y 1 m+n , m z 2 +n z 1 m+n )

x= k⋅( −1 )+1⋅2 k+1 = 2−k k+1

y= k⋅2+1⋅( −3 ) k+1 = 2k−3 k+1

z= k⋅1+1⋅4 k+1 = k+4 k+1

A, B, C are collinear if for some value of k , the points Q and C coincide. Hence,

x=0 2−k k+1 =0 k=2

Substitute k=2 in the values of y and z ,

y= 2k−3 k+1 = 2⋅2−3 2+1 = 4−3 3 = 1 3

z= k+4 k+1 = 2+4 2+1 = 6 3 =2

Thus, C=( 0, 1 3 ,2 ) is a point which divides AB in the ratio 2:1 and is the same as Q. Hence, A, B and C are collinear.