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

Given:

A point $P$ divides the line segment joining the points $A (3, -5)$ and $B (-4, 8)$ such that $\frac{AP}{PB} = \frac{k}{1}$.

$P$ lies on the line $x + y = 0$.

To do:

We have to find the value of $k$.

Solution:

Let the coordinates of point $P$ be $(x,y)$.

If a line segment joining two points two points $(x_{1} ,\ y_{1})$ and $( x_{2} ,\ y_{2} )$ is divided by point $P( x,\ y)$ in the ratio m:n, then,

$P( x,\ y) =\left(\frac{nx_{1} +mx_{2}}{m+n} ,\ \frac{ny_{1} +my_{2}}{m+n}\right)$

Here,

$m=k,\ n=1,\ x_{1} =3,\ y_{1} =-5,\ x_{2} =-4\ and\ y_{2} =8$

Using the division formula, we have, 

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

$\Rightarrow x=\frac{-4k+3}{k+1} \ ,\ y=\frac{8k-5}{k+1}$...............(i)

The point P lies on the line $x+y=0$,

On substituting (i) in the line $x+y=0$, we get,

$\frac{-4k+3}{k+1} +\frac{8k-5}{k+1} =0$

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

$\Rightarrow 4k-2=0$

$\Rightarrow 4k=2$

$\Rightarrow k=\frac{2}{4}$

$\Rightarrow k= \frac{1}{2}$

The value of $k$ is $\frac{1}{2}$.

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Updated on: 10-Oct-2022

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