Here we go,
MENTAL NOTE: Any Line with Positive Slope goes through sectors I and III. And any Line with Negative Slope goes through sectors II and IV. - This is a shortcut here, and you should always remember these crucial bullet points. Hence, from 1, we get slope of K is negative, i.e. -1/6 thus keep (A)
From 2 we get that y-intercept is -6, but we do not get any additional information. For example it could be line k as follow:
y=a*x-6, y=0 --->>>a*x=6, what is a? it could be >0 and x>0 or a<0 and x<0. Thus 2 is insufficient and we are left with (A).

Please, correct me if I went awry

pratik.rossonero wrote:Can anyone explain how to solve this problem? Thanks

A line with -ve slope y= -x/6 + C ; depending on value of C it can lie in II,I and IV quadrants or II,III or IV quadrants and If C=0 then it lies in II and IV quadrants.
Therefore in any case it cuts QII therefore 1. is suficient
Now a line y= m*x -6 may or may not cut QII depending upon the value of m e.g y= x -6 don't cut QII but y = -x -6 cuts the QII hence 2. is not sufficient
A(ans)