Suppose i have points ( 0.9, 79, 76,23, 1, 3, 4.3,89), what is the slope of the line formed by this points

slope of line formed by arbitary points

[Image Analyst]

Those aren't "points" - it's only the x or only the y values. Please supply actual points.

Answer 1

[Azzi Abdelmalek]

These points don't form one line. Maybe you need to use a curve fitting toolbox to fit a function ax+b

[John D'Errico]

ax+b IS the equation of a straight line. How will the curve fitting toolbox fit a straight line any better than any other fitting tool? Magic?

[Azzi Abdelmalek]

Jhon, I have not said it's a better way, just proposed a solution with (maybe). what is yours?

Answer 2

[Wayne King]

You have not told us what the "x"-values are. If we assume that they are

 y = [0.9, 79, 76,23, 1, 3, 4.3,89];
 x = 1:length(y);  % 1 to 8

You can only fit a least-squares line to this data in order to measure the slope.

 X = ones(length(y),2);
 X(:,2) = 1:length(y);
 y = y(:);
 X\y

The intercept is 34.4071 and the slope is 0.0262, but that is a very poor approximation to the data.

Answer 3

[Jos (10584)]

help polyfit

Note that points are usually defined by coordinates and not by single numbers ...

[Azzi Abdelmalek]

If coordinates are not given, we suppose they are equally spaced.

[Jos (10584)]

why?

Answer 4

[Image Analyst]

Try polyfit() to fit (regress) a line through some (x,y) coordinates:

coefficients = polyfit(x, y, 1);
slope = coefficients(1);