Points $P$ and $Q$ trisect the line segment joining the points $A(-2,0)$ and $B(0,8)$ such that $P$ is near to $A$. Find the coordinates of $P$ and $Q$.

Given:

Points $P$ and $Q$ trisect the line segment joining the points $A(-2,0)$ and $B(0,8)$ such that $P$ is near to $A$.

To do:

We have to find the coordinates of $P$ and $Q$.

Solution:

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

$AP: PB = 1:2$ and $AQ:QB=2:1$

Using section formula, we have

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

Then, co-ordinates of $P$ are,

$P=( \frac{1\times0+2\times(-2)}{1+2}, \frac{1\times8+2\times0}{1+2})$

$\Rightarrow P=( \frac{0-4}{3}, \frac{8+0}{3})$

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

Then, co-ordinates of $Q$ are,

$Q=( \frac{2\times0+1\times(-2)}{1+2}, \frac{2\times8+1\times0}{1+2})$

$\Rightarrow Q=( \frac{0-2}{3}, \frac{16+0}{3})$

$\Rightarrow Q=( \frac{-2}{3}, \frac{16}{3})$

The coordinates of $P$ and $Q$ are $(\frac{-4}{3}, \frac{8}{3})$ and $(\frac{-2}{3}, \frac{16}{3})$ respectively.