slope

[ketkoag]

Line L pass point (a, b) and (c,d), if slope <0
1) (a-c)*(b-d)<0
2) Product of interceptions on axis-x and axis-y >0

Here, please tell me how second statement is suff. Also by first statement I took some values and checked, the slope is coming to be positive. will it always be true?

[SanjeevK]

[spoiler]IMO D:[/spoiler]

Since L passes through (a,b) and (c,d), the slope of line is (d-b)/(c-a) or (b-d)/(a-c)

A: (a-c)*(b-d)<0. This means either (a-c) or (b-d) has to be negative but not both. Hence slope will be negative. Sufficient
B: The equation of the line will be y = [(b-d)/(a-c)] x + k (where k is a constant)
The y intercept will be (0,k) and x intercept will be -k(a-c)/(b-d)
Product of x intercept and y intercept is positive.
=> [-k(a-c)/(b-d)]k >0
=> -(k^2)(a-c)/(b-d) > 0. This means (a-c)/(b-d) < 0 . This means slope is negative. Sufficient

Hope this helps

[yogami]

SLope is always (y1 - y2)/(x1 - x2) which is (d - b)/(c - a). We want to prove if this ratio is < 0
(1) says the product is negative. If two numbers when multiplied are negative they will be negative when divided too, so slope is also negative hence sufficient
(2) if products of x and y intercepts are positive this implies those intercepts are both positive or both negative. Now visualize the line. LIne is cutting the x and y axis at positive x and positive y axis somewhere bending towards left like this \. And any line like that has -ve slope. That line can also be cutting -ve x axis and -ve y axis too and still have its shape \. hence sufficient

so D