The slope of the ratio of two lines

[StockholmEd]

Hi, folks!

I need to check that given two lines both of whose slope is negative, then dividing one line by the other yields a third line whose slope is positive. I tried tinkering around with the algebra, but I can't seem to prove it.

Basically given two lines, y1 = m1x + c1, and y2 = m2x + c2, and given that m1 and m2 are both < 0, I thought it would be easy to show that if I divide the first line by the second, then the result would itself be a line whose slope is > 0.

But I can't seem to work it out.

I also tried selecting two points on both lines, via m = (y2 - y1)/(x2 - x1) and simply dividing the coordinates and then trying to create the line formula, but I can't separate the slope of the resulting equations into a form that depends only on x (it keeps getting mixed-up with y and the the c1, c2 constants).

Am I trying something that makes no sense?

Thanks,

Ed

 

[Phrzby Phil]

What do you mean by "dividing one line by the other" ?

If you mean forming the rational function (m1x + c1)/(m2x + c2), then in general you will not get a linear function, and therefore not a straight line, and therefore not a slope.

Example: graph (-x+5) / (-2x+7)

(BTW - why do you use c1 and c2 when conventional terminology is b1 and b2 ?)

 

[StockholmEd]

Oh, thank you! I hadn't thought of just plugging it into some graphing software.

Very interesting.

(And I use c1/c2 because that's what was in maths books 40 years ago; had no idea b1/b2 was the convention now.)

 

[Phrzby Phil]

So did you mean (m1x + c1)/(m2x + c2) ?

Just curious - what is the actual problem, or was this just a curiosity - also a legitimate pursuit.

Also, only books I've ever seen say y = mx+b, where m = slope and b = y-intercept.

 

[StockholmEd]

Yes indeed: I did mean (m1x + c1)/(m2x + c2).

I'm a programmer and trying to calculate the probability of a function's being updated based on its size.

"Also, only books I've ever seen say y = mx+b, where m = slope and b = y-intercept."- Ah. I'm sure I'm misremembering then. It's been an age since I last read this stuff.

 

[apprentus]

Your intuition is correct - dividing two lines with negative slopes should result in a line with a positive slope. Here's one way to see it:

If y1 = m1x + c1 and y2 = m2x + c2, then the line obtained by dividing y1 by y2 is:

y1/y2 = (m1x + c1)/(m2x + c2)

Multiplying both sides by (m2x + c2) gives:

y1 = (m1/m2)x(m2x + c2) + c1(m2x + c2)/y2

y1 = [(m1/m2)x + c1(m2/m2)] + [c2(m1/m2)x + c1c2/y2]

y1 = [(m1/m2)x + c1] + [(c2/y2)(m1x + c1)]

Notice that the first term in the expression in brackets has a positive coefficient (since m1 and m2 are both negative), while the second term has a non-negative coefficient, since c2 and y2 are both positive. Therefore, the slope of the resulting line is positive, since the coefficient of x is positive.

So, dividing two lines with negative slopes results in a line with a positive slope.

 

[StockholmEd]

Thanks for your reply.

In the end, I differentiated to get the slope of the ratio of the lines. See 15:03 here: