The correct option is B II quadrant
Given: x=2,y=−1,u=3,v=2
Putting the values of x,y,u,v in (x×y,u×v)
⟹(2×(−1),3×2)=(−2,6)
We know that if abscissa < 0 and ordinate > 0, then the coordinates will lie in II quadrant.
Since, the value of (x×y)=−2<0 and (u×v)=6>0, (x×y,u×v)will lie in II quadrant.