Find the area of the square $ABCD$ in terms of $u$ and $v$.

Question

QUESTION: Given a square $ABCD$ with two consecutive vertices, say $A$ and $B$ on the positive $x$-axis and positive $y$-axis respectively. Suppose the other vertex $C$ lying in the first quadrant has coordinates $(u , v)$. Then find the area of the square $ABCD$ in terms of $u$ and $v$.

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MY APPROACH: I was trying to solve it out using complex numbers, but I need a minor help. I have assumed $A$ to be $(x_1+0i)$, $B$ to be $(0+y_2i)$ and $C$ is $(u+vi)$. We know that multiplying a point by $i$ basically rotates it by $90°$, about the origin. Here, $C$ is nothing but the reflection of $A$ about the line $BD$. So if I can somehow rotate $A$ about $B$ by $90°$ then we will get $x_1$ and $y_2$ in terms of $u$ and $v$. This is where I am stuck. How to rotate a point with respect to another?

Note that this question has been asked before. But I want to know how to solve it using complex numbers..

Any answers, possibly with a diagram will be much helpful..

Thank you so much..

Answer 1

You can think of changing your frame of reference so that you're rotating around $B$. Look at the vector $BA = x - yi$. Then $BC = (BA)i = y + xi$.

In other words, $C = B + BC = (yi) + (y + xi) = y + (x+y)i$. Therefore $u = y$ and $v = x+y$.

You can calculate the area of the square in terms of $x$ and $y$, then convert that to $u$ and $v$.

Answer 2

You can calculate it like this:

$$(u-v)^2+u^2$$

To understand why, lets define two variables: $x, y$. Given the fact that the square is "leaning" on the y-axis, $x$ is the distance from the origin to the lowest vertex of the square that touches the y-axis and $y$ is the distance from the origin to the furthest vertex of the square that touches the x-axis. Now, we can see that $(u, v)$ is just $(y, x+y)$. Identically, we can see the fourth vertex is at $(x+y, x)$ or $(v, u-v)$. The distance between these is a side length of the square so this distance squared is the area of the square. Using the distance formula we get:

$A = \left(\sqrt{(u-v)^2+(v-(u-v))^2}\right)^2 = \left(\sqrt{(u-v)^2+v^2} \right)^2 = (u-v)^2+u^2$

Answer 3

The area you want is the difference between the large square and 4 triangles. If $u\ge v$, you have another case.

Answer 4

Let $a,b $ be x and y intercepts. Draw lines parallel to x-axis and y-axis.

$$ a= v- u,\; b=u $$

The diagram will be helpful.

$$ Area = a^2+b^2= 2 u^2 +v^2 - 2 u v $$