# Point A lies on the line segment PQ joining P(6, −6) and Q(−4, −1) in such a way that

Question:

Point $A$ lies on the line segment $P Q$ joining $P(6,-6)$ and $Q(-4,-1)$ in such a way that $\frac{P A}{P Q}=\frac{2}{5}$.

If the point $A$ also lies on the line $3 x+k(y+1)=0$, find the value of $k$.

Solution:

Let the coordinates of $A$ be $(x, y)$. Here, $\frac{P A}{P Q}=\frac{2}{5}$. So,

$P A+A Q=P Q$

$\Rightarrow P A+A Q=\frac{5 P A}{2} \quad\left[\because P A=\frac{2}{5} P Q\right]$

$\Rightarrow A Q=\frac{5 P A}{2}-P A$

$\Rightarrow \frac{A Q}{P A}=\frac{3}{2}$

$\Rightarrow \frac{P A}{A Q}=\frac{2}{3}$

Let (xy) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get

$x=\frac{2 \times(-4)+3 \times(6)}{2+3}=\frac{-8+18}{5}=\frac{10}{5}=2$

$y=\frac{2 \times(-1)+3 \times(-6)}{2+3}=\frac{-2-18}{5}=\frac{-20}{5}=-4$

Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore

$3 \times 2+k(-4+1)=0$

$\Rightarrow 3 k=6$

$\Rightarrow k=\frac{6}{3}=2$

Hence, k = 2.