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Straight line passes through the points (a,b) and (c,d). Is [#permalink]

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28 Aug 2009, 06:17

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Straight line passes through the points (a,b) and (c,d). Is the slope of the line is less than 0? (1) (a-c)(b-d)<0 (2) Product of the intercepts of the line on the X-axis and on the Y-axis is greater than 0

I could prove that the 1st statement is enough, but struggling with secone one.

Basically, the question is asking (d-b)/(c-a)<0 i.e. (d-b) and (c-a) have opposite signs 1) (a-c)(b-d)<0, this is same as (c-a)(d-b) <0. Therefore, they have different signs and hence SUFF

2) Product of x-intercept and y-intercept <0

eqn of line with two pts (x1,y1) & (x2,y2) is y-y1 = m(x-x1) substituting m = ((d-b)/(c-a)), we get

y = mx-mx1+y1, where y1-mx1 is the y-intercept and x-intercept = mx1-y1/m

i tried plugging in the values but not getting the solution.

Two points are (a,b) and (c,d) 1) Slope of the line = (d-b)/(c-a) Given (d-b)(c-a) <0, so even the slope < 0. Hence statement 1 alone is sufficient.

2) Given product of x intercept and y intercept is greater than 0. let the points at which the line meets X and Y axis are (x1, 0) and (0, y1) So slope of line = - y1/x1 Given x1*y1 >0 So slope of the line will be negative. Hence statement 2 alone is sufficient.

Statement 1. (a-c)(b-d)<0. 1) a-c>0, then b-d<0 2) a-c<0, then b-d>0 It means that when x increases y decreses=> the line is sloping down or mathematically, the slope \(\frac{a-c}{b-d}<0\) since numerator and denominator have opposite signs. SUFFICIENT

Statement 2 is sufficient If the product of intercepts is positive, intercepts should be of the same sign => this is only the case when a line is sloping down. If the line is sloping "up" the product of the intercepts is negative or zero. Note: if it is said that the product is zero, we can't determine if the line is loping up or down since both are possible. SUFFICIENT. Mathematically it can be also shown: \(y=\alpha\times x +\beta\) Point (0,\(\beta\)) is the Y intercept Point \((- \frac{\beta}{\alpha},0)\)is the X intercept \(-\frac{\beta}{\alpha}\times \beta=-\frac{\beta^2}{\alpha}>0\)...it is only possible when \alpha<0...