A mildly relevant question

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How do you do the equation of a line?

y = mx + b
13
76%
y-y1 = m(x-x1)
4
24%
 
Total votes: 17

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Xisiqomelir
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A mildly relevant question

Post by Xisiqomelir »

I know everyone ends up with y = mx + b, but which do you normally start with?
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Post by Mitth`raw`nuruodo »

I use y=mx+b because it's just.. easier, I guess. Unless you dont know the y-intercept, in which case you have to go with the 2nd one... I think.

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Post by Oddysseus »

I am an old fan of y = mx +b.

But for some higher level statistics and regressions I generally turn and use the alternative. You just need 2 point to determine the slope, then take that and one point to generate the y-value.
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Post by TheDarkOne »

How about Ax + By = C?

or maybe

R=(x,y,z) + n(x',y',z')?
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Post by El Moose Monstero »

I've always quite liked the latter, even though the former was more use, the second one did me well in the Pure Maths exams at AS level.
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Post by Luke Starkiller »

The second is useful for interpolation on steam tables.
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Post by Joe »

y = F + vx

F = Fixed Costs
vx = Variable costs.

This is the equation used in cost accounting to describe cost behavior, and the one I most often use.
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Post by Sharp-kun »

I use y = mx + c
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Post by Sarevok »

Me too.
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Post by haas mark »

y = mx + b was the first I learned. Then progressed to ax + cy = b after that. -shrugs-
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Post by Howedar »

y = ax + b is more convenient, but I more often have to start with y - y1 = m(x - x1)
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Post by SyntaxVorlon »

Bah, always use parametric symmetry.
(x-a)/A=(y-b)/B=(z-c)/C=...
Works in any euclidean geometry.
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Post by 2000AD »

I got taught y = mx + c (or in this case y = mx + b)
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