Given the $4$ starting $x,y$ coordinates of a rectangle and the angle by which it is rotated how to calculate if a point is inside the rectangle I have a starting rectangle with edges parallel to the axis and its center at the origin.  I have all the $x,y$ coordinates of each vertex of the rectangle.  Then I rotate the rectangle by an angle around the origin which is the center of the rectangle. I then have an $x,y$ point which I want to check if it is within the rectangle or outside of it.
Basically to put it another way is there an equation or inequality equation that tells if a given $x$ and $y$ are within the rotated rectangle.
 A: Since rotation won't change the shape of the rectangle all points inside the initial rectangle remain in the rotated rectangle. Meaning if a point is inside the rectangle initially then it will be in the rectangle upon rotation. 
So to check if a point $(x,y)$ is inside the rotated rectangle rotate the point in the opposite direction by the same amount and check if it is in the non rotated rectangle. 
Say the initial rectangle has corners $(x_1, y_1)$ and $(x_2, y_2)$. Now we rotate the by angle $\theta$. Then we have rotation matrix $$\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos\theta\end{bmatrix}$$
For a point $(x,y)$ check if  $$\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos\theta\end{bmatrix}^{-1}\begin{pmatrix}x \\y\end{pmatrix} = \begin{pmatrix}x' \\y'\end{pmatrix}$$
Check $\min(x_1,x_2) \le x'\le\max(x_1,x_2)$ and $\min(y_1,y_2) \le y'\le\max(y_1,y_2)$
$$\begin{pmatrix}x' \\y'\end{pmatrix}=\begin{pmatrix}x\cos\theta-y\sin\theta \\ x\cos\theta+y\sin\theta\end{pmatrix}$$
A: A more general way,
though much less efficient in this case
than Piyush Divyanakar's solution,
will work for
any convex polygon.
The basic subproblem
is decide if
two points are on
the same side of a line.
If the equation of the line is
$ax+by+c = 0$,
substitute the coordinates
of each point
into the equation of the line.
If the results have the same sign,
the points are on the same side of the line;
if not,
they are on different sides.
If you have a convex polygon
(this works in any number of dimensions),
the average of all the vertices
is inside the polygon.
Therefore,
to test if a point 
is inside the polygon,
first
get the equations of
all the bounding lines,
get the center as the average of the vertices,
and get the sign when
the center is substituted
into each bounding line.
Then,
to test the point,
substitute it in
the equation of each bounding line.
If the sign differs from
the sign when the center is substituted,
the point is outside;
if the sign is the same
for all bounding lines,
then the point is inside.
Note: There can be
problems when
the point is close to or on
a bounding line, especially
when the computation is done
in floating point.
