# Find the points of trisection of the line segment joining the points:$(3, -2)$ and $(-3, -4)$<

Given:

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

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 $A (3, -2)$ and $B (-3,-4)$ 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,

$(x,y)=\left[\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right]$
$C(x_1,y_1)=\frac{1 \times(-3)+2 \times 3}{1+2}, \frac{1 \times (-4)+2 \times(-2)}{1+2}$
$=\left(\frac{-3+6}{3}, \frac{-4-4}{3}\right)$
$=\left(\frac{3}{3}, \frac{-8}{3}\right)$

$=\left(1, \frac{-8}{3}\right)$
$D$ intersects $A B$ in the ratio $2: 1$

This implies,

$A D: D B=2: 1$
$D(x_2,y_2)=\left(\frac{(2 \times(-3))+1 \times 3}{2+1}, \frac{2 \times (-4)+1 \times(-2)}{2+1}\right)$

$=\left(\frac{-6+3}{3}, \frac{-8-2}{3}\right)$

$=\left(\frac{-3}{3}, \frac{-10}{3}\right)$

$=\left(-1, \frac{-10}{3}\right)$

The points of trisection of the given segment are $(1, \frac{-8}{3})$ and $(-1, \frac{-10}{3})$.

Updated on: 10-Oct-2022

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