Determining if an arbitrary point lies inside a triangle defined by three points?

Yes, this problem certainly has come up a lot before. Eric Haines wrote about the more general "point in polygon" problem (whose algorithms can be vastly simplified for the triangular case) in http://books.google.com/books?id=CCqzMm_-WucC&pg=PA24. He also presented a method using barycentric coordinates in this http://drdobbs.com/article/print?articleId=184404201.

I think you could check by computing the area of $\triangle ABC$, $\triangle ABP$, $\triangle BCP$, and $\triangle ACP$ (which can be done relatively easily from the coordinates of the points), and if $k_{\triangle ABC}=k_{\triangle ABP}+k_{\triangle BCP}+k_{\triangle ACP}$ (where $k_\_$ is the area), then $P$ is inside the triangle (or possibly on its boundary), but if $k_{\triangle ABC}< k_{\triangle ABP}+k_{\triangle BCP}+k_{\triangle ACP}$, $P$ is outside the triangle.

Method To test any ppoint $P=(x,y)$, first move the origin at one vertex, like $A$ such that $$ B \rightarrow B - A $$ $$ C \rightarrow C - A $$ $$ P \rightarrow P - A $$ Then I calculate the scalar $ d = x_B y_C - x_C y_B $ and the three Barycentric weights $$ \begin{matrix} w_A = \frac{ x ( y_B-y_C) + y ( x_C-x_B) + x_B\; y_C - x_C\; y_B}{d} \\ w_B = \frac{ x\; y_C - y\; x_C }{d} \\ w_C = \frac{ y\; x_B - x\; y_B }{d} \end{matrix} $$ The point is inside when all weights are between $0\ldots1$. Examples Example: $A=(1,1)$ $B=(4,2)$ $C=(2,7)$. Consider a point $P=(2,3)$ then the sclar is $d=17$ and three weights are: $w_A = \frac{8}{17}$, $w_B = \frac{4}{17}$ and $ w_C=\frac{5}{17}$ which are all $|w_i|<1$. On the other hand if $P=(1.5,5)$ then $w_A = \frac{13}{34} $, $w_B = -\frac{1}{17}$ and $ w_C=\frac{23}{34}$ and since $w_B$ does not fall between $0\ldots1$ then the point is outside. Proof Use homogeneous coordinates with $A=(x_A,y_A,1)$, $B=(x_B,y_B,1)$, $C=(x_C,y_C,1)$, $P=(x,y,1)$ and use the following relation $$ P = w_A\;A + w_B\;B+w_C\;C $$ to solve for $w_A$, $w_B$ and $w_C$. The notice that the equation $w_A=0$ described the line $BC$ and the equation $w_A=1$ a line parallel to $BC$ through $A$. Similarly for the other weights. The region where all the weights are $w_i\geq0$ and $ w_i\leq1$ is the triangle described by $ABC$.