Statement 1
Draw the graph of the given equation

the point (x,y) can lie in II,III or IV quadrant

Insufficient

Statement 2
xy>0

this means x and y have the same sign...
Insufficient
so the pont can lie in I or III quadrant

Combining 1 and 2
Point must lie in quadrant



Option C

st(1) y=-x-1 slope is -1 and we conclude that the point (x,y) may potentially be in any quadrant (except for I) Not Sufficient
st(2) xy>0
case a) x>0, y>0 => y=x slope +1 OR II quadrant
case b) x<0, y<0 => y=x slope -1 OR IV quadrant
Not Sufficient

combined st(1&2): When y is +ve, x is +ve too, BUT we have "-" sign in front of x turns it -ve, hence our(x,y) point is II quadrant. This is not possible as y must be negative at all times and (-ve)+(-ve)=(-ve) and not (+ve) as it's required by st(2) y>0. When y is negative, x is -ve too, BUT "-" sign in front of x turn it +ve, hence our (x,y) point is in III quadrant. Sufficient to answer (x,y) can be ONLY in III quadrant.

c

colakumarfanta wrote:Q. (x, y)lies in which quadrant?
(1) x + y =- 1 (2) xy> 0

OA C

del. commment
sorry

Plotting this on graph would make the problem easier. As explained by rijul with

A. x,y can lie in II, III, IV
B. both are positive and both are negative.

Using both we can say it lies on quad III

IMO C.

Using 2: only quad-1 and quad-3 stands (both +ve or both -ve) => -x * -y or x* y both >0
Using 1: only quad-3 stands as in quad-1 all values are positive, so their sum is never negitive.

colakumarfanta wrote:Q. (x, y)lies in which quadrant?
(1) x + y =- 1 (2) xy> 0

OA C