Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Represent the following set of points in the XY plane :

$$\{ ( x , y ) \; | \; |x| + |y| = 1 \}$$

What i got:

1) if $x > 0, y > 0 : x = 1 - y$

2) if $x > 0, y < 0 : x = 1 + y$

3) if $x < 0, y > 0 : x = y - 1$

4) if $x < 0, y < 0 : x = -y -1$

Any help to solve this question would be greatly appreciated.

Thank you,

share|improve this question
    
What you have works. You can check this by noting that if x or y<0, then -x or -y is positive, and thus you can use -x or -y for |x| or |y| in the original equation. If x or y>0, then use x or y in the original equation. What's your question exactly? –  Doug Spoonwood Jun 10 '12 at 10:20
add comment

1 Answer 1

up vote 2 down vote accepted

Take each case without the absolute value:$$\,\,\,y=1-x$$$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y=1-(-x)=1+x$$$$-y=1-x$$$$-y=1+x$$so you can see we get four straight lines intersecting each with other two.

Well, now just draw these lines and get your nice... rhomboid (you could know this even before drawing anything from the above equations).

share|improve this answer
    
Rhomboid? It's a square! –  Rahul Jun 10 '12 at 11:10
    
Well @Rahul, you know: by definition, any square is a rhomboid...and a rectangle as well. I chose to call it rhomboid because of its relative position wrt both axis... –  DonAntonio Jun 10 '12 at 12:12
    
Well, I figure it's better to be more precise than less so. But also, according to Wikipedia and MathWorld, a square actually isn't a rhomboid by definition. –  Rahul Jun 10 '12 at 12:20
    
This seems to be a matter of terminology/language. I've no problem in calling the figure a rhombus and not a rhomboid if most authors do agree on this, but the fact is that a parallelogram with all its sides of equal length is a rhombus and thus a square is a rhombus. It doesn't really matter, I think, for this particular question. –  DonAntonio Jun 10 '12 at 13:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.