Question

what can we generalize about a sqaure with two unknown sides

Answer 1

Please post diagram if there is one.

Answer 2

Try this :
\[\begin{align}
B&= \left(\dfrac{a+b+c-d}{2}, ~~\dfrac{-a+b+c+d}{2}\right)\\~\\~\\
D&= \left(\dfrac{a-b+c+d}{2}, ~~\dfrac{a+b-c+d}{2}\right)\\~\\~\\

\end{align}\]

Answer 3

For area simply find the length of side by dividing the diogonal by \(\sqrt{2}\), then square it :

\[s = \dfrac{\sqrt{(a-c)^2 + (b-d)^2}}{\sqrt{2}}\]
\[\implies A = s^2 = \dfrac{(a-c)^2 + (b-d)^2}{2}\]

Answer 4

can u explain the logic to the equations? @ganeshie8 thank u

Answer 5

familiar with rotation matrix ?
http://en.wikipedia.org/wiki/Rotation_matrix

Answer 6

thats okay.. linear algebra is not really required here :)

Answer 7

do you know how to rotate a point around origin by an angle ?

Answer 8

:( no

Answer 9

lets see if we can work it using distance formula
Let \(P(x,y)\) be the point that stays at a distance of \(\dfrac{\overline{AC}}{\sqrt{2}}\) from both vertices \(A(a,b)\) and \(C(c,d)\) :

\[(x-a)^2 + (y-b)^2 = \frac{1}{2}[(a-c)^2 + (b-d)^2]\tag{1}\]
\[(x-c)^2 + (y-d)^2 = \frac{1}{2}[(a-c)^2 + (b-d)^2]\tag{2}\]

Solve both the equations simultaneously for \(x,y\). You will get two solution pairs representing the points \(B\) and \(D\)

Answer 10

where do u get square root of 2 from?

Answer 11

and I I don't understand your eqautions either.

im hopeless