Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
\(\left(2, -\dfrac{5}{3}\right)\) and \(\left(0, -\dfrac{7}{3}\right)\)
Let the endpoints of the line segment be \(A(4, -1)\) and \(B(-2, -3)\).
The points of trisection divide the line segment \(AB\) into three equal parts. Hence, they divide the segment internally in the ratios \(1:2\) and \(2:1\).
Using the section formula, a point dividing the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) internally in the ratio \(m:n\) has coordinates:
\( \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \)
First trisection point (ratio \(1:2\))
Here, \(m = 1\), \(n = 2\), \((x_1, y_1) = (4, -1)\), \((x_2, y_2) = (-2, -3)\).
\(x = \frac{1(-2) + 2(4)}{3} = \frac{-2 + 8}{3} = 2\)
\(y = \frac{1(-3) + 2(-1)}{3} = \frac{-3 - 2}{3} = -\frac{5}{3}\)
So, the first trisection point is \(\left(2, -\frac{5}{3}\right)\).
Second trisection point (ratio \(2:1\))
Here, \(m = 2\), \(n = 1\), \((x_1, y_1) = (4, -1)\), \((x_2, y_2) = (-2, -3)\).
\(x = \frac{2(-2) + 1(4)}{3} = \frac{-4 + 4}{3} = 0\)
\(y = \frac{2(-3) + 1(-1)}{3} = \frac{-6 - 1}{3} = -\frac{7}{3}\)
Thus, the second trisection point is \(\left(0, -\frac{7}{3}\right)\).
Therefore, the points of trisection of the given line segment are \(\left(2, -\dfrac{5}{3}\right)\) and \(\left(0, -\dfrac{7}{3}\right)\).