Find the area of the square using co-ordinates Given a square $ABCD$ such that the vertex $A$ is on the $x$-axis and the vertex $B$ is on the $y$-axis. The coordinates of vertex $C$ are $(u,v)$. Find the area of square in terms of $u$ and $v$ only.
What I have done
Let the coordinate of $A$ be $(x,0)$ and $B$ be $(0,y)$. Also let the side of the square be a units.
$2a^2=AC^2=(x-u)^2+v^2$
$a^2$ is the required area so if we write $x$ in terms of $u$ and $v$ then the job will be done. 
Now from here I thought of two ways either using trigonometry or using rotation of axes but here none of them will work because some angles will be involved and we require just $u$ and $v$ and nothing else in the expression of the area of square.
So how to do it? Please help.

Let $A(x;0), B(0,y), C(u,v)$, and $AB=a$. Then $$S=a^2$$.
$$\begin{cases}
AB=a=\sqrt{x^2+y^2}
\\
BC=a=\sqrt{u^2+(v-y)^2}{}
\\
AC=\sqrt2a=\sqrt{(u-x)^2+v^2}
\end{cases}$$
$$\begin{cases}
a^2=x^2+y^2 (1)
\\
a^2=u^2+(v-y)^2 (2)
\\
2a^2=(u-x)^2+v^2 (3)
\end{cases}$$
(3)+(2) $3a^2=u^2-2ux+x^2+v^2+u^2+v^2-2vy+y^2=2u^2-2ux+2v^2-2vy+(x^2+y^2)=$
$=2u^2-2ux+2v^2-2vy+a^2 \Rightarrow a^2=u^2-ux+v^2-vy$
$$\begin{cases}
a^2=x^2+y^2 (1)
\\
a^2=u^2-ux+v^2-vy (4)
\end{cases}$$
