Question

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 $A _{1}$ and $A _{2}$ be the areas of $\Delta ABC$ and $\Delta PQC$ respectively, such that $A _{1}=3 A _{2}$, then the value of $m$ is equal to :

• A. $\frac{4}{15}$
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$P \equiv\left(x_{1}, m x_{1}\right)$
$Q \equiv\left(x_{2}, m x_{2}\right)$
$A_{1}=\frac{1}{2} \begin{vmatrix} 3 & 4 & 1 \\ 2 & 0 & 1 \\ -1 & 1 & 1\end{vmatrix} -\frac{13}{2}$
$A_{2}=\frac{1}{2} \begin{vmatrix} x_{1} & m x_{1} & 1 \\ x_{2} & m x_{2} & 1 \\ 2 & 0 & 1\end{vmatrix}$
$A_{2}=\frac{1}{2}\left|2\left(m x_{1}-m x_{2}\right)\right|-m\left|x_{1}-x_{2}\right|$
$A_{1}=3 A_{2} \Rightarrow \frac{13}{2}=3 m\left|x_{1}-x_{2}\right|$
$\Rightarrow \left|x_{1}-x_{2}\right|-\frac{16}{6 m}$
$A C: x+3 y=2$
$B C: y=4 x-8$
$P: x+3 y=2\,\& y\,=m x \Rightarrow x_{1}=\frac{2}{1+3\,m}$
$Q: y=4 x-8 \,\& \,y=m x \Rightarrow x_{2}=\frac{8}{4-m}$
$\left|x_{1}-x_{2}\right|-\left|\frac{2}{1+3\,m}-\frac{8}{4-m}\right|$
$\left|\frac{-26\,m}{(1+3 m)(4-m)}\right|=\frac{26\,m}{(3\,m+1)|m-4|}$
$=\frac{26\,m}{(3\,m+1)(4-m)}$
$\left|x_{1}-x_{2}\right|=\frac{13}{6\,m}$
$\frac{26 m}{(3\,m+1)(4-m)}=\frac{13}{6 m}$
$\Rightarrow 12\,m^{2}=-(3\,m+1)(m-4)$
$\Rightarrow 12\,m^{2}=-\left(3\,m^{2}-11 m-4\right)$
$\Rightarrow 15\,m^{2}-11\,m-4=0$
$\Rightarrow 15\,m^{2}-15\,m+4 m-4=0$
$\Rightarrow (15\,m+4)(m-1)=0$
$\Rightarrow m=1$