Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$.

Given: 

Given points are $(4, -1)$ and $(-2, -3)$.

To do: 

We have to find the points of trisection of the line segment joining the given points.

Solution:

Let the line segment whose end points are $A (4, -1)$ and $B (-2,-3)$ is trisected at points $C(x_1,y_1)$ and $D(x_2,y_2)$.
$C$ divides the line segment in the ratio $1 : 2$

This implies,

$AC : CB = 1 : 2$

Therefore,

Using the division formula

$(x,y)=[\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}]$

$C(x_1,y_1)=\frac{1 \times(-2)+2 \times 4}{1+2}, \frac{1 \times (-3)+2 \times(-1)}{1+2}$

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

$=(\frac{6}{3}, \frac{-5}{3})$

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

$D$ intersects $AB$ in the ratio $2: 1$

This implies,

$A D: D B=2: 1$

$D(x_2,y_2)=(\frac{(2 \times(-2))+1 \times 4}{2+1}, \frac{2 \times (-3)+1 \times(-1)}{2+1})$

$=(\frac{-4+4}{3}, \frac{-6-1}{3})$

$=(\frac{0}{3}, \frac{-7}{3})$

$=(0, \frac{-7}{3})$

The points of trisection of the given segment are $(2, \frac{-5}{3})$ and $(0, \frac{-7}{3})$.