Since, the x-coordinate a of point R is less than 0 and the y-coordinate b of point R is greater than 0, it means the point R lies in the 2nd quadrant where the x-axis is negative and y-axis is positive.
x-coordinate of point R < 0 implies a < 0
y-coordinate of point R > 0 implies b > 0
Similarly, x-coordinate m and y-coordinate n of the point S are greater than 0, it means the point R lies in the 1st quadrant where both axis are positive.
x-coordinate of point S > 0 implies m > 0
y-coordinate of point S > 0 implies n > 0
However, we are also told that the y-coordinate b of the point R is greater than the y-coordinate n of the point S.
y-coordinate of point R > y-coordinate of point S
a > m
After understanding the placement of both points $R$ and $S$, we can say that the line will go downwards because the point R will be placed higher in the 2nd quadrant as compared to the point S in the 1st quadrant. Therefore, the slope of the line \(\bar{RS}\) will be negative.
Quantity A = Slope of \(l_{1}\), which is negative
Quantity B = 0
Quantity A < Quantity B
Hence, the correct answer is Option B).
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