Find the area of the square using co-ordinates

Question

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.

Answer 1

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) $3a^2=u^2-2ux+x^2+v^2+u^2+v^2-2vy+y^2=2u^2-2ux+2v^2-2vy+(x^2+y^2)=$ $=2u^2-2ux+2v^2-2vy+a^2 \Rightarrow a^2=u^2-ux+v^2-vy$

$$\begin{cases} a^2=x^2+y^2 (1) \\ a^2=u^2-ux+v^2-vy (4) \end{cases}$$

Answer 2

If the points $A(x,0)$ and $B(0,y)$ are adjacent vertices of a square, then the square’s area is $x^2+y^2$. Assuming that the vertices are enumerated counterclockwise, $C$ has coordinates $(-y,y-x)$, i.e., $B+\operatorname{rot}(B-A,\pi/2)$, so $x=-(u+v)$, $y=-u$ and $x^2+y^2=(u+v)^2+u^2=2u^2+2uv+v^2$. If the vertices are enumerated clockwise instead, then $C$ is $(y,x+y)$, $x=v-u$, $y=u$ and the area is $2u^2-2uv+v^2$.