Find coordinates of the points of trisection of the line segment joining the point.

A (-4,2) and B (-4, 9)

$\begin{matrix} P = (- 4, \frac{13}{3}) Q = (- 4, \frac{20}{3}) \end{matrix}$

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$\begin{matrix} P = (- 4, \frac{20}{3}) Q = (- 4, \frac{13}{3}) \end{matrix}$

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$\begin{matrix} P = (- 4, 6) Q = (- 4, 8) \end{matrix}$

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$\begin{matrix} P = (4, 9) Q = (9, \frac{7}{4}) \end{matrix}$

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Solution

The correct option is A
$\begin{matrix} P = (- 4, \frac{13}{3}) Q = (- 4, \frac{20}{3}) \end{matrix}$

For point P, $m_{1} : m_{2} = A P : P B = 1 : 2$

$(x_{1}, y_{1}) = (- 4, 2) a n d (x_{2}, y_{2}) = (- 4, 9)$

$∴ x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} = \frac{(- 4) + 2 (- 4)}{3}$

$= \frac{- 4 - 8}{3} = \frac{- 12}{3} = - 4$

$∴ y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} = \frac{1 (9) + 2 (2)}{3} = \frac{13}{3}$

$∴$ point $P = (- 4, \frac{13}{3})$

For $Q, m_{1} : m_{2} = A Q : Q B = 2 : 1 : (x_{1}, y_{1}) = (- 4, 2)$ and $(x_{2}, y_{2}) = (- 4, 9)$

$∴ x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} = [\frac{2 (- 4) + 1 (- 4)}{3}] = \frac{- 12}{3}$

$y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} = [\frac{2 (9) + 1 (2)}{3}] = \frac{20}{3}$

$∴ Q = (- 4, \frac{20}{3})$

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