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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 $(-1, \frac{-10}{3})$.
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