As you know the equation of a line is y=kx+b
II quadrant means that for x<0 y>0
(1) says that the line is y=-1/6x+b. Whatever b you have (positive or negative), this is some finite number. Taking x as low as you can (going to -infinity), you always could get -1/6*x as much positive as you need such that the sum -1/6x+b is positive. In other words, solving -1/6*x+b=0 you get the for any x<6b y is positive, obviously for some negative values of x this will also hold.
(2) says that y(0)=-6 which means that b=-6. Is the slope is positive, then such line will never intersect II quadrant. For example y=-6+x will never intersect it.
So (1) is sufficient, (2) is not. The answer is (A)
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