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

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Solution

Let P (x₁, y₁) and Q (x₂, y₂) are the points of trisection of the line segment joining the given points.
$\Rightarrow$ AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2 so using section formula.
$x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} a n d y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}$
In this case $(m_{1} = 1, m_{2} = 2 a n d x_{1} = 4, x_{2} = - 2, y_{1} = - 1, y_{2} =_{3})$
$x_{1} = \frac{1 \times (- 2) + 2 \times 4}{1 + 2}, y_{1} = \frac{1 \times (- 3) + 2 \times (- 1)}{1 + 2}$
$x_{1} = \frac{- 2 + 8}{3} = \frac{6}{3} = 2, y_{1} = \frac{- 3 - 2}{3} = \frac{- 5}{3}$
Therefore,
$P (x_{1}, y_{1}) = (2, \frac{- 5}{3})$
Point Q divides AB internally in the ratio 2:1
$x_{2} = \frac{2 \times (- 2) + 1 \times 4}{2 + 1}, y_{1} = \frac{2 \times (- 3) + 1 \times (- 1)}{2 + 1}$
$x_{2} = \frac{- 4 + 4}{3} = 0, y_{2} = \frac{- 6 - 1}{3} = \frac{- 7}{3}$
$P (x_{2}, y_{2}) = (0, \frac{- 7}{3})$

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

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