Question

A point P divides the line segment joining the points A(3, −5) and B(−4, 8) such that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{k}{1}.$ If P lies on the line x + y = 0, then find the value of k. [CBSE 2012]

Answer 1

It is given that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{k}{1}.$

So, P divides the line segment joining the points A(3, −5) and B(−4, 8) in the ratio k : 1.

Using the section formula, we get

Coordinates of P = $\left(\frac{-4k+3}{k+1},\frac{8k-5}{k+1}\right)$

Since P lies on the line x + y = 0, so

$$\begin{gathered} \frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0 \\ \Rightarrow \frac{-4k+3+8k-5}{k+1}=0 \\ \Rightarrow 4k-2=0 \\ \Rightarrow k=\frac{1}{2} \end{gathered}$$

Hence, the value of k is $\frac{1}{2}$.