Answer :

Let P (x_{1}, y_{1}) and Q (x_{2}, y_{2}) 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.

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

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

\(\therefore \)P = (x_{1}, y_{1}) = P(2, \(\frac{-5}{3} \))

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

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

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

The coordinates of the point Q(x_{2} ,y_{2} ) = (0, \(\frac{-7}{3} \) )

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