Point P divides the line segment joining the points $\displaystyle A( 2,1)$ and $\displaystyle B( 5,-8)$ such that $\displaystyle \frac{AP}{AB} =\frac{1}{3}$. If P lies on the line $\displaystyle 2x-y+k=0$, find the value of k.

Given: A line segment joins two points $A( 2,1)$ and $B( 5,-8)$.There is a point P lying on the given line segment such that $\frac{AP}{AB} =\frac{1}{3}$ and P also lies on the line $2x-y+k=0$.

To do: To find the value of k.

Solution:
First of all we draw a line AB and locate point P on line segment AB.
Here given 
$A\ ( 2,\ 1) ,\ point\ B\ ( 5,\ -8) .$

$P\ lies\ on\ the\ line\ 2x-y+k=0$

$Let\ us\ say\ P\ ( x,\ y)$

First we will find the ratio of division,

$Here\ P\ divides\ the\ line\ segment\ such\ that\ \frac{AP}{AB} =\frac{1}{3}$

$\therefore AP=\frac{AB}{3}$

$\because AB=AP+BP$

$\Rightarrow BP=AB-AP$

$\Rightarrow BP=AB-\frac{AB}{3}=\frac{3AB-AB}{3}=\frac{2}{3}AB$

$\therefore \frac{AP}{BP} =\frac{\frac{AB}{3}}{\frac{2AB}{3}}$

$\Rightarrow \frac{AP}{BP} =\frac{1}{2}$

$\Rightarrow AP∶BP=1:2$

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

$P( x,\ y) \ =(\frac{2\times 2-5\times 1}{2+1} ,\frac{2\times 1-8\times 1}{2+1} )$

$=(\frac{4-5}{3} ,\ \frac{2-8}{3} )$

$=\ ( 3,\ -2)$

By putting the value of $P( 3,\ -2)$ in the line $2x-y+k=0$
$2( 3) -2+k=0$

$\Rightarrow k=-4$

Thus the value of $k=4$.
 

Updated on: 10-Oct-2022

42 Views

Related Articles

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements