Prove the following

Solution:

$\mathrm{P} \equiv\left(\mathrm{x}_{1}, \mathrm{~m} \mathrm{x}_{1}\right)$

$\mathrm{Q} \equiv\left(\mathrm{x}_{2}, \mathrm{mx}_{2}\right)$

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

$\mathrm{A}_{2}=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{x}_{1} & \mathrm{mx}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{mx}_{2} & 1 \\ 2 & 0 & 1\end{array}\right|$

$\mathrm{A}_{2}=\frac{1}{2}\left|2\left(\mathrm{mx}_{1}-\mathrm{mx}_{2}\right)\right|=\mathrm{m}\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|$

$\mathrm{A}_{1}=3 \mathrm{~A}_{2} \Rightarrow \frac{13}{2}=3 \mathrm{~m}\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|$

$\mathrm{AC}: \mathrm{x}+3 \mathrm{y}=2$

$B C: y=4 x-8$

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

$\mathrm{Q}: \mathrm{y}=4 \mathrm{x}-8 \& \mathrm{y}=\mathrm{mx} \Rightarrow \mathrm{x}_{2}=\frac{8}{4-\mathrm{m}}$

$\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|=\left|\frac{2}{1+3 \mathrm{~m}}-\frac{8}{4-\mathrm{m}}\right|$

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

$=\frac{26 \mathrm{~m}}{(3 \mathrm{~m}+1)(4-\mathrm{m})}$

$\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|=\frac{13}{6 \mathrm{~m}}$

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

$\Rightarrow 12 \mathrm{~m}^{2}=-(3 \mathrm{~m}+1)(\mathrm{m}-4)$

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

$\Rightarrow 15 \mathrm{~m}^{2}-11 \mathrm{~m}-4=0$

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

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

$\Rightarrow \mathrm{m}=1$