Let $A (- 1, 1), B (3, 4) a n d C (2, 0)$ be given three points. A line $y = m x, m > 0,$ intersects lines AC and BC at point P and Q respectively. Let $A_{1}$ and $A_{2}$ be the areas of $Δ A B C$ and $Δ P Q C$ respectively, such that $A_{1} = 3 A_{2},$ then the value of $m$ is equal to :

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Detailed Solution

$A_{1} = Δ A B C = \frac{1}{2} ||\begin{matrix} - 1 & 1 & 1 \\ 2 & 0 & 1 \\ 3 & 4 & 1 \end{matrix}|| = \frac{1}{2} (- 1 (- 4) - 1 (- 1) + 1 (8)) = \frac{13}{2} square units$

$\begin{array}{l} Equation of the line A C is y - 1 = \frac{1}{3} (x + 1) \\ Solve it with the line y = m x, we get P (\frac{2}{3 m + 1}, \frac{2}{3 m + 1}) \\ Equation of line B C is y = 4 (x - 2) \\ Solve it with the line y = m x we get Q (\frac{8}{m - 4}, \frac{8 m}{m - 4}) \end{array}$

Area of $Δ P Q C = \frac{1}{2} ||\begin{matrix} 2 & 0 & 1 \\ \frac{2}{3 m + 1} & \frac{2 m}{3 m + 1} & 1 \\ \frac{- 8}{m - 4} & \frac{- 8 m}{m - 4} & 1 \end{matrix}|| = \frac{A_{1}}{3} = \frac{13}{6} \Rightarrow \frac{1}{2} m |\begin{matrix} 2 & 0 & 1 \\ \frac{2}{3 m + 1} & \frac{2}{3 m + 1} & 1 \\ - \frac{8}{m - 4} & \frac{- 8}{m - 4} & 1 \end{matrix}| = \frac{13}{6} \Rightarrow \frac{m}{2} |\begin{matrix} 2 & 0 & 1 \\ 0 & \frac{2}{3 m + 1} & 1 \\ 0 & - \frac{8}{m - 4} & 1 \end{matrix}| = \frac{13}{6} \Rightarrow m (\frac{2}{3 m + 1} + \frac{8}{m - 4}) = \frac{13}{6}$

$\begin{matrix} S i m p l i f y i n g t h e a b o v e e q u a t i o n \\ \frac{26 m^{2}}{(3 m + 1) (m - 4)} = \pm \frac{13}{6} \\ 12 m^{2} = \pm (3 m^{2} - 11 m - 4) \\ 9 m^{2} + 11 m + 4 = 0 o r \\ 15 m^{2} - 11 m - 4 = 0 \end{matrix}$