The line segment joining the points $A( 2,\ 1)$ and $B( 5,\ -8)$ is tri-sected at the points P and Q such that P is nearer to A. If P also lies on the line given by $2x - y = 0$, find the value of $k$.

Given: The line segment joining the points $A( 2,\ 1)$ and $B( 5,\ -8)$ is trisected at the points $P$ and $Q$ such that $P$ is nearer to $A$ and $P$ also lies on the line given by $2x - y = 0$

To do: To find the value of k.

Solution:

As line segment $AB$ is trisected by the points $P$ and $Q$. 

When $AP: PB = 1:2$

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 $P$ are:

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

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


$\Rightarrow P=( \frac{9}{3}, \frac{-6}{3})$

$\Rightarrow P=( 3, -2)$

As given the point $P( 3,\ -2)$ lies on the line $2x-y+k=0$ 

Then point $P( 3,\ -2)$ will satisfy the equation,

$\Rightarrow 2(3)-(-2)+k=0$

$\Rightarrow 6+2+k=0$

$\Rightarrow 8+k=0$

$\Rightarrow k=-8$

Therefore, $k=-8$

 

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Updated on: 10-Oct-2022

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