Position of Point with respect to a Triangle Consider a triangle formed by three sides, equations of which are as follows - 
a1x + b1y + c1=0
a2x + b2y + c2=0
a3x + b3y + c3=0
Is there a nice method to figure out whether a point P(h,k) lies inside or outside the triangle formed by these lines? 
It'd be great if some sort of a formula could be derived using the variables mentioned above. In fact, a brief outline of the approach followed while getting to the formula would be great too. 
Thanks a lot. 
 A: Here's a method (not a formula) . . .


*

*Solve each pair of two equations in two unknowns to find the vertices, $U,V,W$ say.$\\[4pt]$

*Check to make sure that $U,V,W$ are not collinear.$\\[4pt]$

*Find the centroid $G={\large{\frac{U+V+W}{3}}}$.$\\[4pt]$

*Let $p_1,p_2,p_3$ and $g_1,g_2,g_3$ be the results of plugging the coordinates of $P$ and $G$, respectively, into the left-hand-sides of each of the three equations.


Then


*

*If at least one of the products $p_1g_1,p_2g_2,p_3g_3$ is negative, then $P$ is outside triangle $UVW$.$\\[4pt]$

*If all three of the products $p_1g_1,p_2g_2,p_3g_3$ are nonnegative, and at least one of them is zero, then $P$ is on the boundary of triangle $UVW$.$\\[4pt]$

*If all three of the products $p_1g_1,p_2g_2,p_3g_3$ are positive, then $P$ is inside triangle $UVW$.

A: Method:
1) Solve the equations pairwise to find the vertices $A,B,C$ of the triangle $ABC$.
2) Calculate the areas of $\Delta APB$, $\Delta BPC$, $\Delta APC$ and $\Delta ABC$.
3) If $S_{\Delta ABC}=S_{\Delta APB}+S_{\Delta BPC}+S_{\Delta APC}$, then $P$ is inside. If LHS is less than RHS, then $P$ is outside.
