Interior point of $\Delta\,ABC$ if $P(\lambda,2)$ is an interior point of $\Delta\,ABC$ formed by the lines
$$x+y=4$$ $$3x-7y=8$$ $$4x-y=31$$ Find $\lambda$
My Idea: The vertices of $\Delta ABC$ are 
$A(\frac{18}{5},\frac{2}{5})$ $\:$ $B(7,-3)$ and $C(\frac{209}{25},\frac{61}{25}) $ 
So using areas of $\Delta ABC$, $\Delta APC$,$\Delta ABP$,$\Delta PBC$ we have
$$\begin{vmatrix}
 &\frac{18}{5}  & \frac{2}{5} & 1\\ 
 & 7 & {-3}& 1\\ 
 &\frac{209}{25}  &\frac{61}{25} &1 
\end{vmatrix}=\begin{vmatrix}
 &\frac{18}{5}  & \frac{2}{5} & 1\\ 
 & \lambda & {2}& 1\\ 
 &\frac{209}{25}  &\frac{61}{25} &1 
\end{vmatrix}+\begin{vmatrix}
 &\frac{18}{5}  & \frac{2}{5} & 1\\ 
 & 7 & {-3}& 1\\ 
 &\lambda  &2 &1 
\end{vmatrix}+\begin{vmatrix}
 &\lambda  & 2 & 1\\ 
 & 7 & {-3}& 1\\ 
 &\frac{209}{25}  &\frac{61}{25} &1 
\end{vmatrix}$$ from which we get $\lambda$
Is there any smarter way to do this..

Another way is rough plot all these three lines with $y=2$ With seeing the points of intersection and a rough idea of their positions you will be able to directly tell the range of $\lambda$ 
