# A point P divides the line segment joining the points

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 $\frac{1}{2}$.