Question

How to derive the equation of a square that is |x+y|+|x-y|=a?

Answer 1

This might help
http://www.derivative-calculator.net/

Answer 2

Not finding the derivative...

Answer 3

Like how do you know that the equation of a square is |x+y|+|x-y|=1

Answer 4

sorry then

this mod will help.

Answer 5

lolz i know this mod

Answer 6

and he just said he can't help me

Answer 7

You know she is so stupid that she can't help, yea :\

Answer 8

lolz

Answer 9

the question is ambiguous

Answer 10

Like I know the equation of a square is \(|x+y|+|x-y|=a\), but how

Answer 11

I'm not sure exactly if this is what you want...
Uhmmm.. Like how can we graph a square given that particular function?
Is that what you want to know?

Answer 12

I already know how to express the equation of a square (|x+y|+|x-y|=a), I just wanna know the working principle :)

Answer 13

http://www.wolframalpha.com/input/?i=ContourPlot++Abs%5Bx+-+y%5D+%2B+Abs%5Bx+%2B+y%5D+

Answer 14

You argue like this if \( \large x \ge y>0 \), then
\[
\large (|x+y|+|x-y|=a\\
x+y + x -y =a=2x\\
x=\frac a 2
\]
and you consider all other cases, you find that
\[
x=\pm \frac a 2\\
y=\pm \frac a 2
\]

Answer 15

These are the equations of the sides of a square of side a and centered at the origin.

Answer 16

It sorta reminds me of the standard form for a circle. (AND I SAY SORTA!)

Answer 17

|x+y|+|x-y|=a
i might have an idea but dnt know if it gonna work

Answer 18

but the final sketch i guss it goes like this one

Answer 19

@ikram002p

\[
2 \sqrt{(x-y)^2}\sqrt{(x+y)^2}= 2 |x-y| |x+y|

\] and not


\[
2 \sqrt{(x-y)^2}\sqrt{(x+y)^2}= 2 (x-y)(x+y)
\]

Answer 20

@ikram002p also the final graph looks like
http://www.wolframalpha.com/input/?i=ContourPlot++Abs%5Bx+-+y%5D+%2B+Abs%5Bx+%2B+y%5D+

one of the above

Answer 21

the final i drowed was for |2x|=|a|
humm i know there is somthing wrong but not completly
the analys i give is one of the cases cuz as u said
2bla bla=2|x+y||x-y|
so i can covert it to 2(x+y)(x-y)
cuz at the end i can elimate y
but the idea is how to convert a squire
cus
|2x|=|a|
is not the same of
|x+y|+|x-y|=a
its just a case of it

Answer 22

Anytime you are dealing with absolute value of a number you have to consider cases where the number inside is positive or negative.
\[\sqrt{(-5)^2} = 5=|-5|\quad and\, not -5
\]

Answer 23

\[
\sqrt{x^2}= |x|
\]

Answer 24

You have to do it the way I suggested above by considering all the possibilities

Answer 25

ohk maby next time :)
ty for ur advice

Answer 26

Your method @ikram002p is not 100% correct that is what I am trying to explain to you.

Answer 27

i know i said that from the beginig i said this befor i answer

"i might have an idea but dnt know if it gonna work"
i was trying to solve it for real

Answer 28

One also can easily prove the opposite, that is if you take any point ( u,v) on the lines
\[
x=\pm \frac a 2 \\
y=\pm \frac a 2
\]
then
|u-v| + |u+v|=a
Also you have to consider 8 cases ( all are similar) and 2 are in each quadrant.

Answer 29

Of course restricted to the boundary of the square.

Answer 30

Let us consider on case, the other are similar
case 1. First quadrant
\[
u=\frac a 2, \, 0 \leq v \leq \frac a 2 \\
|u-v| + |u+v| = \left | \frac a 2 -v \right|+ \left | \frac a 2+v \right|=\\
\frac a 2 -v + \frac a 2+v=a

\]

Answer 31

one case

Answer 32

thanks so much everyone :D

Acknowledgement: @GABI_J @Callisto @alekos @zepdrix @eliassaab @myininaya @ikram002p

Answer 33

equation of a square ... that was new for me lol

Answer 34

That's why I'm trying to understand it lol :)

Answer 35

if we consider it a surface: z = f(x,y) .... but then it could also be seen as a square "tube" sliced, hmmm