Question

Using section formula, show that he points A(2, –3, 4), B(–1, 2, 1) and C(0, 1/3, 2) are collinear.

Answer 1

Let, C divides AB in the ratio $\lambda$:1.
Then, coordinates of C are $\left(\frac{-\lambda +2}{\lambda +1},\frac{2\lambda -3}{\lambda +1},\frac{\lambda +4}{\lambda +1}\right)$.
But, the coordinates of C are $\left(0,\frac{1}{3},2\right)$.
∴ $\frac{-\lambda +2}{\lambda +1}$=0 , $\frac{2\lambda -3}{\lambda +1}$=$\frac{1}{3}$, $\frac{\lambda +4}{\lambda +1}$ = 2
From each of these equations, we get $\lambda =2$.
Since each of these equations give the same value of $\lambda$.
Therefore, the given points are collinear.