Question

Point A lies on the line segment PQ joining P(6, −6) and Q(−4, −1) in such a way that $\frac{PA}{PQ}=\frac{2}{5}$.
If the point A also lies on the line 3x + k (y + 1) = 0, find the value of k. [CBSE 2015]

Answer 1

Let the coordinates of A be (x, y). Here, $\frac{PA}{PQ}=\frac{2}{5}$. So,
$$\begin{gathered} PA+AQ=PQ \\ \Rightarrow PA+AQ=\frac{5PA}{2}[\because PA=\frac{2}{5}PQ] \\ \Rightarrow AQ=\frac{5PA}{2}-PA \\ \Rightarrow \frac{AQ}{PA}=\frac{3}{2} \\ \Rightarrow \frac{PA}{AQ}=\frac{2}{3} \end{gathered}$$
Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
$$\begin{gathered} x=\frac{2\times \left(-4\right)+3\times \left(6\right)}{2+3}=\frac{-8+18}{5}=\frac{10}{5}=2 \\ y=\frac{2\times \left(-1\right)+3\times \left(-6\right)}{2+3}=\frac{-2-18}{5}=\frac{-20}{5}=-4 \end{gathered}$$
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
$$\begin{gathered} 3\times 2+k\left(-4+1\right)=0 \\ \Rightarrow 3k=6 \\ \Rightarrow k=\frac{6}{3}=2 \end{gathered}$$
Hence, k = 2.