If ab ≠0 and (-a, b) and (-b, a) are in the same quadrant, is (-x, y) in this
quadrant?
a. xy > 0
b. ax > 0
OA is C
Quadrants
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- sivaelectric
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Can you please Explain Akansha
If I am wrong correct me , If my post helped let me know by clicking the Thanks button .
Chitra Sivasankar Arunagiri
Chitra Sivasankar Arunagiri
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As (-a,b) and (-b,a) are in same quadrant - it means both a and b are of same sign.
If they are of different signs, (assume a>0 and b<0)(-a,b) will be in 3rd quadrant while (-b,a) will be in 1st quadrant.
Thus ab>0
also (-a,b) and (-b,a) are either in 2nd quadrant or 4th quadrant.
a) xy>0 => (-x,y) is either in 2nd or 4th quadrant. But there might be a case when (-a,b) is in 2nd quad while (-x,y) in 4th quad
Or a case when both (-a,b) & (-x,y) be in 4th quad
Thus insufficient.
b) ax>0
suppose if a>0, (-a,b) in 2nd quad
also then x>0 and (-x,y) in 2nd or 3rd quad (we don't know whether y is +ve or -ve)
Thus insufficient.
both together =>
if a>0 => x>0 and y>0 and thus (-a,b) in 2nd quad and also (-x,y) in 2nd quad
if a<0 => x<0 and y<0 and thus (-a,b) in 4th quad and also (-x,y) in 4th quad
Thus both xy>0 and ax>0 => (-a,b), (-b,a) & (-x,y) are in same quadrant.
Hence C
If they are of different signs, (assume a>0 and b<0)(-a,b) will be in 3rd quadrant while (-b,a) will be in 1st quadrant.
Thus ab>0
also (-a,b) and (-b,a) are either in 2nd quadrant or 4th quadrant.
a) xy>0 => (-x,y) is either in 2nd or 4th quadrant. But there might be a case when (-a,b) is in 2nd quad while (-x,y) in 4th quad
Or a case when both (-a,b) & (-x,y) be in 4th quad
Thus insufficient.
b) ax>0
suppose if a>0, (-a,b) in 2nd quad
also then x>0 and (-x,y) in 2nd or 3rd quad (we don't know whether y is +ve or -ve)
Thus insufficient.
both together =>
if a>0 => x>0 and y>0 and thus (-a,b) in 2nd quad and also (-x,y) in 2nd quad
if a<0 => x<0 and y<0 and thus (-a,b) in 4th quad and also (-x,y) in 4th quad
Thus both xy>0 and ax>0 => (-a,b), (-b,a) & (-x,y) are in same quadrant.
Hence C
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