I think its C - both combined are sufficient. I went about this one with a graphical solution, which is kind of hard to explain without drawings, but hopefully this helps. Start by drawing a coordinate plane with the x and y axes. Since we know the line will pass through the origin, you can put a point at (0,0).
(1) gives us that the slope is negative, so our line will slope down from left to right intersecting with the origin. So, it will start in the upper left quadrant of the plane, pass through (0,0) and into the lower right quadrant. If you draw such a line, you will quickly see that the y-coordinate of its points can be either positive (when x-coord is negative) or negative (when x is positive). Therefore, we cannot say whether b is positive or negative.
(2) gives us that a<b. That is, the x-coordinate of the point on our line that we are interested in will be less than the y-coordinate. On your coordinate plane, draw a line through the origin sloping upward from left to right with slope 1...this is the line y=x. Statement (2) tells us that the point (a,b) must be in the region above this line (where y>x). However, you will see that this region includes areas where b could be positive or negative (can be neg. as long as it is less negative the x-coordinate a).
If we combine (1) and (2) however, we know that the point (a,b) is above the line y=x (from 2) and in either the upper left or lower right quadrant (from 1). The only points that satisfy both of these are all in the upper left quadrant and such points all have a y-coordinate that is positive, so we know that b will be >0.
