The line segment joining the points $P (3, 3)$ and $Q (6, -6)$ is trisected at the points $A$ and $B$ such that $A$ is nearer to $P$. If $A$ also lies on the line given by $2x + y + k = 0$, find the value of $k$.

Given:

The line segment joining the points $P (3, 3)$ and $Q (6, -6)$ is trisected at the points $A$ and $B$ such that $A$ is nearer to $P$ and $A$ also lies on the line given by $2x + y + k = 0$.

To do:

We have to find the value of $k$.

Solution:

As line segment $PQ$ is trisected by the points $A$ and $B$. 

$PA: AQ = 1:2$

Using section formula, we have

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

Then, co-ordinates of $A$ are:

$A=( \frac{1\times6+2\times3}{1+2}, \frac{1\times(-6)+2\times3}{1+2})$

$\Rightarrow A=( \frac{6+6}{3}, \frac{-6+6}{3})$

$\Rightarrow A=( \frac{12}{3}, \frac{0}{3})$

$\Rightarrow A=( 4, 0)$

The point $A( 4,\ 0)$ lies on the line $2x+y+k=0$.

This implies, point $A( 4,\ 0)$ satisfies the above equation.

$\Rightarrow 2(4)+0+k=0$

$\Rightarrow 8+k=0$

$\Rightarrow k=-8$

The value of $k$ is $-8$.

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Updated on: 10-Oct-2022

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