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  • Let A (-1, 1), B ( 3,4) and C (2,0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the area of <img alt="\Delta ABC" src="/latex-image/?%5C

Let A (-1, 1), B ( 3,4) and C (2,0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the area of \Delta ABC and \Delta PQC respectively, such that A_1 = 3A_2, then the value of m is equal to:
Option: 1 \frac{4}{15}
Option: 2 3
Option: 3 2
Option: 4 1

Answers (1)

best_answer

 

\begin{array}{l} P \equiv\left( x _{1}, mx _{1}\right) \\ Q \equiv\left( x _{2}, mx _{2}\right) \end{array}

\\A_{1}=\frac{1}{2}\left|\begin{array}{ccc} 3 & 4 & 1 \\ 2 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right|=\frac{13}{2} \\ \\\\A_{2}=\frac{1}{2}\left|\begin{array}{ccc} x_{1} & m x_{1} & 1 \\ x_{2} & m x_{2} & 1 \\ 2 & 0 & 1 \end{array}\right|

\\A _{2}=\frac{1}{2}\left|2\left( mx _{1}- mx _{2}\right)\right|= m \left| x _{1}- x _{2}\right| \\ \\A _{1}=3 A _{2} \Rightarrow \frac{13}{2}=3 m \left| x _{1}- x _{2}\right|

\\\Rightarrow \quad\left| x _{1}- x _{2}\right|=\frac{13}{6 m } \\ \\AC : x +3 y =2 \\ \\BC : y =4 x -8

\\P : x +3 y =2 \& y = mx \Rightarrow x _{1}=\frac{2}{1+3 m } \\ \\Q : y =4 x -8 \& y = mx \Rightarrow x _{2}=\frac{8}{4- m }

\left| x _{1}- x _{2}\right|\\\\=\left|\frac{2}{1+3 m }-\frac{8}{4- m }\right| \\ \\\\=\left|\frac{-26 m }{(1+3 m )(4- m )}\right|\\\\=\frac{26 m }{(3 m +1)| m -4|}\\\\=\frac{26 m }{(3 m +1)(4- m )}

\begin{aligned} &\left| x _{1}- x _{2}\right|=\frac{13}{6 m } \\ &\frac{26 m }{(3 m +1)(4- m )}=\frac{13}{6 m } \\ &\Rightarrow 12 m ^{2}=-(3 m +1)( m -4) \\ &\Rightarrow 12 m ^{2}=-\left(3 m ^{2}-11 m -4\right) \\ &\Rightarrow 15 m ^{2}-11 m -4=0 \\ &\Rightarrow \quad 15 m ^{2}-15 m +4 m -4=0 \\ &\Rightarrow(15 m +4)( m -1)=0 \\ &\Rightarrow m =1 \end{array}

 

Posted by

Suraj Bhandari

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