Interior point of $\Delta\,ABC$

Question

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..

Answer 1

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 $\lambda$

Answer 2

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.