If the straight line k does not pass through the origin, then the diagonal equation is y=kx+b(b is not equal to 0), and k is the slope. Of course, the sufficient condition of negative slope in the problem is that the diagonal k is not perpendicular to the X axis.

Option A If he said the intercept on X and Y, it means that the intercept exists, so there is no case that the diagonal is parallel to the X axis or Y axis.

Suppose that the oblique line intersects the X axis at point (a, 0) and the Y axis at point (0, b), where a=2b and b/a =1/2 > 0。 Then the intercept k = (0-b)/(a-0) =- 1/2.

B has ab>0 on the basis of A, that is, K =-B/A.

C actually, (a–r) (b–s) and (b–s)/(a–r), that is, the sign of the slope k is the same. Of course, (a–r) (b–s)