Question

Using section formula show that the points $A(2,-3,4),B(-1,2,1)$ and $C(0,\frac{1}{3},2)$ are collinear.

Answer 1

The given points are $A(2,-3,4),B(-1,2,1)$ and $C(0,\frac{1}{3},2)$
Let $P$ be a point that divides $AB$ in the ratio $k:1$.
Using section formula, the coordinates of $P$ are given by,
$(\frac{k(-1)+2}{k+1},\frac{k\left(2\right)-3}{k+1},\frac{k\left(1\right)+4}{k+1})$
Now, we will find the value of $k$ at which point $P$ coincides with point $C$.
$\Rightarrow \frac{-k+2}{k+1}=0$, we get $k=2$
For $k=2$, the coordinates of point $P$ are $(0,\frac{1}{3},2)$,
i.e., $C(0,\frac{1}{3},2)$ is a point that divides $AB$ internally in the ratio $2:1$ and is the same as point $P$
Hence, points $A,B$ and $C$ are collinear.