Question

If $P(3,2,-4),Q(5,4,-6)$ and $R(9,8,-10)$ are collinear points, then the ratio in which $Q$ divides $PR$, is

• A. $3:2$ (internally)
• B. $3:2$ (externally)
• C. $1:2$ (internally)
• D. $1:2$ (externally)

Answer 1

Answer: (C)

Let $Q$ divides $PR$ in the ratio $k:1$

Here, the point $Q$ divides the line $PR$ internally, so its
coordinates are $\left[\left(\frac{m{x}_2+n{x}_1}{m+n}\right),\left(\frac{m{y}_2+n{y}_1}{m+n}\right),\left(\frac{m{z}_2+n{z}_1}{m+n}\right)\right]$
$Q=\left(\frac{k\times 9+1\times 3}{k+1},\frac{k\times 8+1\times 2}{k+1},\frac{k(-10)+1\times (-4)}{k+1}\right)$
$=\left(\frac{9k+3}{k+1},\frac{8k+2}{k+1},\frac{-10k-4}{k+1}\right)$
But given, $Q=(5,4,-6)$
On comparing the corresponding coordinates
$\therefore \frac{9k+3}{k+1}=5,\frac{8k+2}{k+1}=4,\frac{-10k-4}{k+1}=-6$
$\Rightarrow 9k+3=5k+5,8k+2=4k+4,$
$\Rightarrow 4k-4=-6k-6$
$\Rightarrow 4k=2$
$k=\frac{1}{2}$
Hence, point $Q$ divides $PR$ internally in the ratio $1:2$.