Question:
A point $P$ divides the line segment joining the points $A(3,-5)$ and $B(-4,8)$ such that $\frac{A P}{P B}=\frac{k}{1}$. If $P$ lies on the line $x+y=0$, then find the value of k
Solution:
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 $\mathrm{P}=\left(\frac{-4 k+3}{k+1}, \frac{8 k-5}{k+1}\right)$
Since P lies on the line x + y = 0, so
$\frac{-4 k+3}{k+1}+\frac{8 k-5}{k+1}=0$
$\Rightarrow \frac{-4 k+3+8 k-5}{k+1}=0$
$\Rightarrow 4 k-2=0$
$\Rightarrow k=\frac{1}{2}$
Hence, the value of $k$ is $\frac{1}{2}$.