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
$$\frac{1+(-1)}{2}=\frac{4+x}{2}\quad\text{and}\quad \frac{1+4}{2}=\frac{3+y}{2}.$$
**Hint:** You should be able to draw your parallelogram even if you don't know the coordinates of $D$, at least approximately.

Assuming that you are labelling the parallelogram's vertices counter-clockwise (so that $D$ has negative abscissa), what can you tell about the projection of the side $DA$ on the $y$ axis compared with the projection of the side $AB$? What about the projection of the side $DA$ on the $x$ axis compared with the projection of the side $BC$?

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For example, write $x(P)$ for the $x$ coordinate of a point $P$. Note that you can get to $D$ by "moving" from $A$ along the side $AD$, which you know must be congruent to $BC$. Thus
$$
x(D) = x(A) + x(C) - x(B) = -x(B) = -4.
$$
You can reason similarly about $y(D)$.