Question: ABCD is a parallelogram. A, B and C equal (1,1), (4, 3) and (-1, 4) respectively. 

I have worked out the length of AB and AC to be the root of 13, and ABC is a right-angled triangle as I was asked to prove that (by showing AB and AC are equal). The gradient of AB = 2/3.

I'm simply stuck on finding co-ordinate D. The topic we are doing in class is co-ordinate geometry, so I know about finding the length and straight line equations.

I'd appreciate help or direction in what to do.

Answer: 
Use the fact that the midpoint of the line segment $AC$ is also the midpoint of the line segment $BD$.

Let $D(x,y)$. Then, you have
$$\frac{1+(-1)}{2}=\frac{4+x}{2}\quad\text{and}\quad \frac{1+4}{2}=\frac{3+y}{2}.$$
Assuming $ ABDC $ anti-clockwise, as parallel lines must have equal $x,y$ differences,
$$ x +1 = 4-1, y-4=3-1$$

or

$$ x= 2, y=6$$

$D$ makes side with $B$ for given order; $A$ and $D$ are diagonally opposite. A rough sketch settles the matter with vertex placement order.