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Find the points of trisection of the line segment joining the points:$(2, -2)$ and $(-7, 4)$<
Given:
Given points are $(2, -2)$ and $(-7, 4)$.
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 $A (2, -2)$ and $B (-7,4)$ 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,
\( (x,y)=\left[\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right] \)
\( C(x_1,y_1)=\frac{1 \times(-7)+2 \times 2}{1+2}, \frac{1 \times 4+2 \times(-2)}{1+2} \)
\( =\left(\frac{-7+4}{3}, \frac{4-4}{3}\right) \)
\( =\left(\frac{-3}{3}, \frac{0}{3}\right) \)
\( =\left(-1, 0\right) \)
$D$ intersects \( A B \) in the ratio $2: 1$
This implies,
\( A D: D B=2: 1 \)
\( D(x_2,y_2)=\left(\frac{(2 \times(-7))+1 \times 2}{2+1}, \frac{2 \times 4+1 \times(-2)}{2+1}\right) \)
\( =\left(\frac{-14+2}{3}, \frac{8-2}{3}\right) \)
\( =\left(\frac{-12}{3}, \frac{6}{3}\right) \)
\( =\left(-4, 2\right) \)
The points of trisection of the given segment are $(-1, 0)$ and $(-4, 2)$.
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