Just using coordinates and the clue in the original diagram there is a short and sweet solution sent in by Jo from Leeds.

By concentrating on the geometry, she made the algebra very simple. With the clue that $p-q = a-c$ and $p+q=b-d$ she was able to find the area of the square $ABCD$:

If the vertices were $(6,7)$ and $(3,2)$, the
surrounding square would have side $(7-2)$.

The inner square would have a side of $6-3 =3$.

The area of the tilted square is therefore $5 \times 5 - \frac{(5 \times 5)-(3 \times 3)}{2} =17 $ sq units and this can be generalised.

In the algebraic case.

Area $= (b-d)^2- \frac{((b-d)^2- (a-c)^2)}{2} $

Michael Gray from Madras College solved the first part using vectors.

The midpoint of the square is given by $M=((a+c)/2, (b+d)/2)$. Hence

\begin{eqnarray} \mathbf{CM} &=& \vec{m}
- \vec{c} \\ &=& \frac{1}{2} {a-c \choose b-d}
\end{eqnarray}

The vector $\mathbf{MD}$ is the vector $\mathbf{CM}$ rotated by
$90^o$ anticlockwise and so:
\begin{eqnarray} \mathbf{MD} &=&
\frac{1}{2} {-(b-d) \choose a-c} \\ \mathbf{OD} &=&
\mathbf{OM} + \mathbf{MD} \\ &=& \frac{1}{2} {a+c \choose
b+d} + \frac{1}{2} {-(b-d) \choose a-c} \\ &=& \frac{1}{2}
{a-b+c+d \choose a+b-c+d} \end{eqnarray}

and this gives the coordinates of the point $D$. The vector
$\mathbf{OB}$, giving the coordinates of $B$, is found in a similar
way $$\mathbf{OB} = \frac{1}{2} {a+b+c-d \choose -a+b+c+d}.$$