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

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

x1 = $$\frac{(1×(-2)+2×4)}{1+2} = \frac{(-2+8)}{3} =\frac{6}{3} = 2$$

y1 = $$\frac{(1×(-3)+2×-1)}{1+2} = \frac{(-3-2)}{3} =\frac{-5}{3}$$

$$\therefore$$P = (x1, y1) = P(2, $$\frac{-5}{3}$$)

Point Q divides AB internally in the ratio 2:1.

x2 = $$\frac{(2×(-2)+1×4)}{2+1} = \frac{(-4+4)}{3} = 0$$

y2 = $$\frac{(2×(-3)+1×-1)}{2+1} = \frac{(-6-1)}{3} =\frac{-7}{3}$$

The coordinates of the point Q(x2 ,y2 ) = (0, $$\frac{-7}{3}$$ )