• Aug 31st 2009, 08:15 AM
sharkman
In which quadrant is P(x,y) located if xy < 0 and x + y < 0?

MY WORK:

For xy > 0, I figured that for this to be true, both x and y must be positive values.

In other words, x and y cannot be 0 and negative.

For x + y < 0, one of the variables must be bigger than the other and negative.

I don't know where to go with this question in order to find the quadrant.
• Aug 31st 2009, 09:02 AM
Chris L T521
Quote:

Originally Posted by sharkman
In which quadrant is P(x,y) located if xy < 0 and x + y < 0?

MY WORK:

For xy > 0, I figured that for this to be true, both x and y must be positive values.

In other words, x and y cannot be 0 and negative.

For x + y < 0, one of the variables must be bigger than the other and negative.

I don't know where to go with this question in order to find the quadrant.

Given $a>0,b>0$, What if $x=-a$ and $y=-b$? Then $xy=(-a)(-b)=ab>0$. So x and y can also be negative! (this now tells us P(x,y) is either in the first or third quadrant)

Then it follows that $x+y=(-a)+(-b)=-a-b<0$. This satisfies the conditions of $P\!\left(x,y\right)$ falling into the third quadrant.

Does this make sense?
• Aug 31st 2009, 09:07 PM
sharkman
yes...
Quote:

Originally Posted by Chris L T521
Given $a>0,b>0$, What if $x=-a$ and $y=-b$? Then $xy=(-a)(-b)=ab>0$. So x and y can also be negative! (this now tells us P(x,y) is either in the first or third quadrant)

Then it follows that $x+y=(-a)+(-b)=-a-b<0$. This satisfies the conditions of $P\!\left(x,y\right)$ falling into the third quadrant.

Does this make sense?

Very easily explained.