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..

You can make a rough plot of lines and see whether the point lies above or below the required lines. 

Simply write equation of lines with $y$ coefficient positive. If $L(P)$ is positive, the point is above and otherwise below.e.g. : 
Take $4x-y=31\implies y+31-4x=0$
Now put $(\lambda,2)$
From diagram $P$ should be aove the line,i.e. : $2+31-4\lambda>0$
Obviously, you will, and should, get a range of $\lambda$ unless it lies on one of the lines.
