Question

Given that $P(3,2,-4),Q(5,4,-6)$ and $R(9,8,-10)$ are collinear. Find the ratio in which $Q$ divides $PR$.

Answer 1

Given points $P(3,2,-4),Q(5,4,-6)$ and $R(9,8,-10)$
Let $Q$ divides $PR$ in the ratio $k:1$.
So, by section formula, coordinates of $Q$ are $(\frac{k\left(9\right)+1\left(3\right)}{k+1},\frac{k\left(8\right)+1\left(2\right)}{k+1},\frac{k(-10)+1(-4)}{k+1}$
$=(\frac{9k+3}{k+1},\frac{8k+2}{k+1},\frac{-10k+4}{k+1})$
But the given coordinates of $Q$ are $(5,4,-6)$.
On comparing $\frac{9k+3}{k+1}=5$
$\Rightarrow 9k+3=5k+5$
$\Rightarrow k=\frac{1}{2}$
So, $Q$ divides $PQ$ in the ratio $1:2$.