Let $A\left(-1,1\right), B\left(3,4\right)$ and $\mathrm{C}\left(2,0\right)$ 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 $A_1=3A_2$, then the value of $m$ is equal to :

$P\equiv \left(x_1,m{x}_1\right)$

 $Q\equiv \left(x_2,m{x}_2\right)$

$A_1=\frac{1}{2}\left|\begin{array}{ccc}3& 4& 1\\ 2& 0& 1\\ -1& 1& 1\end{array}\right|=\frac{13}{2}$

$A_2=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& m{x}_1& 1\\ x_2& m{x}_2& 1\\ 2& 0& 1\end{array}\right|$

$A_2=\frac{1}{2}\left|2\left(m{x}_1-m{x}_2\right)\right|=m\left|x_1-x_2\right|$

$A_1=3{ A}_2\Rightarrow \frac{13}{2}=3 m\left|x_1-x_2\right|$

$\Rightarrow \left|x_1-x_2\right|=\frac{16}{6 m}$

$AC: x+3 y=2$

$BC: y=4 x-8$

$P:x+3y=2 \& y=mx\Rightarrow x_1=\frac{2}{1+3m}$

$Q: y=4x-8 \& y=mx\Rightarrow x_2=\frac{8}{4-m}$

$\left|x_1-x_2\right|=\left|\frac{2}{1+3m}-\frac{8}{4-m}\right|$

$=\left|\frac{-26 m}{\left(1+3m\right)\left(4-m\right)}\right|=\frac{26 m}{\left(3m+1\right)\left|m-4\right|}$

$=\frac{26 m}{\left(3m+1\right)\left(4-m\right)}$

$\left|x_1-x_2\right|=\frac{13}{6 m}$

$\frac{26 m}{\left(3m+1\right)\left(4-m\right)}=\frac{13}{6 m}$

$\Rightarrow 12m^2=-\left(3m+1\right)\left(m-4\right)$

$\Rightarrow 12m^2=-\left(3m^2-11m-4\right)$

$\Rightarrow 15m^2-11m-4=0$

$\Rightarrow 15m^2-15m+4m-4=0$

$\Rightarrow \left(15m+4\right)\left(m-1\right)=0$

$\Rightarrow m=1$