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giugio

slope of a line

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giugio    246

hello.
I'm new to trigonometry and math and i'm studing it.
For practice i'm creating a c++ program that gets the points of a line (as input)and calculates the direction or slope of this line(as output).
I'm trying to use the trigonometry but without success because i wish calculate the slope of the line for each point with the tangent.
My key problem is that if i create a vector from the origin to the point and calculate the tangent for each vector of this type i get wrong results, but this  is related to take the origin as start point.
How i can calculate the slope of a line from his points?
That are not ordered.
Thanks.

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zacaj    667

If you're fixed on using trigonometry, you should be able to use atan2(y2-y1,x2-x1) to get the angle of the line, and then plug that into cos and sin to get the slope

 

However, trigonometry doesn't really deal with "slopes".  The slope of a line would be the type used in y=mx+b style formulas.  You can just do m=(y2-y1)/(x2-x1) to get the slope.  

 

What form you actually need depends on what you're doing with the slope 

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Emergent    982

How many points are you requesting as input?  (I understand how two points specify a line.  But if you have three or more points, then I would like to understand how you want to interpret them.  For example, are you drawing a polyline?  Or do you just have two points?)

 

You write,

 

 i wish calculate the slope of the line for each point with the tangent.

 

The slope of a line is the same everywhere; it does not depend on what point you look at.

 

Also, what do you mean by "tangent?"  Do you mean the "tan" function from trigonometry?  I ask because the "tangent" to a curve at a point is a line that (a) passes through that point, and (b) has the same slope as the curve at that point.  But the tangent to a line, at any point, is just the line itself...

 

How i can calculate the slope of a line from his points?

 

If your line contains the points (x1,y1) and (x2, y2), then its slope is,

 

m = (y2 - y1)/(x2 - x1) .

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el junk    191

Sorry if I misinterpret anything you said.

 

What I understood is that you want to study a curve. In math, a curve is more general than a line. For example, it can be "curvy". A line, though, is "plain" and "straight".

 

If your curve happens to be the output of a function (i.e. it passes the vertical line rule), then the ordering of the input points doesn't matter, in that you can first sort them according to their x-coordinate. Otherwise, you would need additional input to tell you how to parameterize the curve (i.e. the order of the input points).

 

For a curve, there exists a tangent line for every point on the curve. For a line (again, a line is a curve), the tangent lines happen to all be coincident.

 

The input points are discrete, but a curve is continuous. Given a finite set of points, infinitely many curves can pass through them. There is no "the curve" for these points. However, if your points meet certain criteria and you impose certain constraints, then there can be such a thing (such as the interpolating polynomial of a degree no more than one less than that of the number of points). One way to make "smooth" curves is to impose constraints on a higher order derivatives of an interpolating function.

 

A simple starting point could be to assign each point the average slope from its neighboring lines. For example:

Input: p0, p1, p2, p3, ..., pn

 

(for 0 <= t <= 1)

L1(t) = p0 + t * (p1 - p0)

L2(t) = p1 + t * (p2 - p0)

...

Ln(t) = p{n-1} + t * (pn - p{n-1})

 

slope(p0) = (1/1) * (slope(L1)) 

slope(p1) = (1/2) * (slope(L1) + slope(L2))

slope(p2) = (1/2) * (slope(L2) + slope(L3))

...

slope(pn) = (1/1) * (slope(Ln))

 

As Emergent mentioned:

slope(L1) = (p1:y - p0:y) / (p1:x - p0:x)

...

 

In my opinion, paper is superior to computer for learning mathematics, but learning how to program mathematics is worth the effort. When making your calculations, check things like divide by zero and look into the implications of using floating point numbers, instead of infinite precision real numbers.

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