Defining set of interior points of a triangle Is there a way, given that $z_1,z_2 \ \text{and} \ z_3$ are the vertices of a triangle in the complex plane, to characterize all point that are inside of the triangle?
 A: Any convex combination of the vertices will give you the points in a triangle.
i.e., $\sum_{k=1}^3 z_k c_k$ for $c_k \in [0,1]$ and $\sum_{k=1}^3 c_k=1$
If you would like the set of interior points only, let $c_k \in (0,1)$ in the above.
In general, the same is true of any convex polygon, since the convex combinations of points generate the convex hull of these points, which is what a polygon really is.
A: What I have done in the past
is to first
compute the center of the triangle
as
$C =(z_1+z_2+z_3)/3
=(x_c, y_c)
$.
Then,
for any pair of points
$z_i$ and
$z_j$,
the line connecting them,
$L_{i, j}$,
can be gotten from
the standard two-point form
of the line
in the form
$L_{i, j}(x, y)
=a_{i,j}x+b_{i,j}y+c_{i,j}
=0
$.
Any scale factor can be used.
A standard one is
to impose
$a_{i,j}^2+b_{i,j}^2
=1$
(though this may not always
be possible).
For any point
$(x, y)$,
the value of
$L_{i,j}(x, y)$
determines the side
of $L_{i,j}$
that $(x, y)$
is on.
This is a scaled and signed version
of the distance
from the point to the line.
Compute
which side of each line
the center $C$ is
by
$s_{i,j} =L_{i,j}(x_c, y_c)
$.
For a non-degenerate triangle
(the three points
are not on the same line),
$s_{i,j}$
will not be zero.
Note that
the value of
$s_{i,j}$
does not matter,
only its sign.
Then,
an arbitrary point
$(x, y)$
is inside the triangle
if and only if
$L_{i,j}(x, y)$
has the same sign
($< 0, 0,$ or $> 0$)
as
$s_{i,j}
$.
Note that
$L_{i,j}(x,y)
=0
$
means that
$(x,y)$
is on 
$L_{i,j}$.
You can interpret this
as inside or outside
as you choose.
This works for
any convex polygon,
not just triangles.
It also works in more than two dimensions.
As in almost everything
in computational geometry,
equations which appear nice
(such as these)
may misbehave for
certain input parameters
(such as three points
almost on the same line).
