help

|x|+|y|=1
when x and y are mutually perpendicular lines or axes
explain

5 Answers

1357
Manish Shankar ·

first quadrant x+y=1
second quadrant x-y=1
first quadrant -x-y=1
fourth quadrant -x+y=1

This will give a rhombus(square) passing thru (0,1),(1,0),(0,-1),(-1,0)

1
preeti kochar ·

why u take 1 everytime in R.H.S

1357
Manish Shankar ·

you are right.. in a way only..

suppose we get an equation

x-y=-1

then we can bring it back to the form -(x-y)=-(-1)

that will give y-x=1

you have to see which equation will be releant in which quadrant...

Here it so happens that we are able to get all these equations such that rhs is -1!

1
ith_power ·

you missed a point manish!!

in the equation rhs is always positive. but x & y can be +ve or -ve.

62
Lokesh Verma ·

Well power u are right and wrong :D

It was only to explain to preeti why the RHS was always +1

I think it is not a "Correct" explanation for people who know exactly what is going on.. But sometimes this way helps.

Of course, you are right in saying that

This equation is

± x ± y = 1

in the appropriate quadrants :)

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