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 $P R$, 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 $P R$ 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{9 k+3}{k+1}, \frac{8 k+2}{k+1}, \frac{-10 k-4}{k+1}\right)$
But given, $Q=(5,4,-6)$
On comparing the corresponding coordinates
$\therefore \frac{9 k+3}{k+1}=5, \frac{8 k+2}{k+1}=4, \frac{-10 k-4}{k+1}=-6$
$ \Rightarrow 9 k+3=5 k+5,8 k+2 =4 k+4, $
$\Rightarrow 4 k-4 =-6 k-6$
$ \Rightarrow 4 k =2 $
$k =\frac{1}{2}$
Hence, point $Q$ divides $P R$ internally in the ratio $1: 2$.