Finding the equation of a line passing 2 points

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Soroush Asarzadeh
Soroush Asarzadeh 2016-7-16
评论： Image Analyst 2023-1-29
Hello, I have two points (x1,y1) and (x2,y2). Now I want to find the linear equation of a line passing through these 2 points. The equation must be like f(x)=a*x+b. Is there any function in matlab that accepts coordinates of two points an gives the related linear equation back? If not, I know that a=(y2-y1)/(x2-x1) but what is the short and easy way to find 'b'? Thanks in advance!
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Ahmed
Ahmed 2021-12-27
You can select Polynomial of degree 1 from the cftool
this will give you something like this
Linear model Poly1:
f(x) = p1*x + p2
Coefficients:
p1 = xx
p2 = xx
where xx could be any number

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采纳的回答

Image Analyst
Image Analyst 2016-7-16
Try polyfit:
coefficients = polyfit([x1, x2], [y1, y2], 1);
a = coefficients (1);
b = coefficients (2);
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Jose
Jose 2022-11-27
When I use polyfit, It tells me that "x1" does not exist.
Image Analyst
Image Analyst 2022-11-27
@Jose then define it. Certainly you must know what points you fit the line through. So, what variable names did you use? Use those instead of x1, etc.

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更多回答（3 个）

Star Strider
Star Strider 2016-7-16
A third (and probably the most efficient) option for this particular problem:
x = [1 2];
y = [5 4];
c = [[1; 1] x(:)]\y(:); % Calculate Parameter Vector
slope_m = c(2)
intercept_b = c(1)
slope_m =
-1
intercept_b =
6
This uses the mldivide,\ operator to do a least-squares fit of the points.
Alok Mishra
Alok Mishra 2022-9-15
编辑：Alok Mishra 2022-9-15 
    function [a b c eq] = makeEquationFrom2Points(x1,y1,x2,y2)
        syms x y;
        if x1==x2 && y1==y2
            disp('Need 2 distinct points');
            a=NaN;
            b=NaN;
            c=NaN;
            eq="null";
            return;
        else if x1==x2
            b=0;
            a=1;
            c=x1;
        else
            %for lines m not_equal to inf 
            %y=mx+c
            %m is coeff(1) and coeff(2) is c 
            coefficient=polyfit([x1 x2],[y1 y2],1);
            a=-coefficient(1);
            c=coefficient(2);
            b=1;
        end
        eq=a*x+b*y==c;
    end

on workspace do:

      [a b c eq]=makeEquationFrom2Points(1,2,3,4)

Output: a =

   -1.0000

b =

     1

c =

    1.0000

eq =

y - x == 1

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navid seif
navid seif 2023-1-29
Thanks.
May I ask how do it works when x1,x2,y1and y2 are vector?
Image Analyst
Image Analyst 2023-1-29
@navid seif put the function in a loop over x1
for k = 1 : numel(x1)
if x1(k) == x2(k) && y1(k) == y2(k)
% etc.
end
end
Everything should have an index k, like a(k), b(k), etc., so that you store the results for every set of data.

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Azzi Abdelmalek
Azzi Abdelmalek 2016-7-16
After you found a, You can get b from your equation y=a*x+b,

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