# Represent a given set of points by the best possible straight line

Find the value of m and c such that a straight line y = mx + c, best represents the equation of a given set of points (x, y ), (x, y ), (x, y ), ……., (x, y ), given n >=2.

Examples:

Input : n = 5 x = 1, x = 2, x = 3, x = 4, x = 5 y = 14, y = 27, y = 40, y = 55, y = 68 Output : m = 13.6 c = 0 If we take any pair of number ( x, y ) from the given data, these value of m and c should make it best fit into the equation for a straight line, y = mx + c. Take x = 1 and y = 14, then using values of m and c from the output, and putting it in the following equation, y = mx + c, L.H.S.: y = 14, R.H.S: mx + c = 13.6 x 1 + 0 = 13.6 So, they are approximately equal. Now, take x = 3 and y = 40, L.H.S.: y = 40, R.H.S: mx + c = 13.6 x 3 + 0 = 40.8 So, they are also approximately equal, and so on for all other values. Input : n = 6 x = 1, x = 2, x = 3, x = 4, x = 5, x = 6 y = 1200, y = 900, y = 600, y = 200, y = 110, y = 50 Output : m = -243.42 c = 1361.97

**Approach**

To best fit a set of points in an equation for a straight line, we need to find the value of two variables, m and c. Now, since there are 2 unknown variables and depending upon the value of n, two cases are possible –

**Case 1 – When n = 2 : **There will be two equations and two unknown variables to find, so, there will be a unique solution .

**Case 2 – When n > 2 : **In this case, there may or may not exist values of m and c, which satisfy all the n equations, but we can find the best possible values of m and c which can fit a straight line in the given points .

So, if we have n different pairs of x and y, then, we can form n no. of equations from them for a straight line, as follows

f = mx + c, f = mx + c, f = mx + c, ......................................, ......................................, f = mx + c, where, f, is the value obtained by putting x in equation mx + c.

Then, since ideally f should be same as y, but still we can find the f closest to y in all the cases, if we take a new quantity, U = ?(y – f ), and make this quantity minimum for all value of i from 1 to n.

**Note:**(y – f ) is used in place of (y – f ), as we want to consider both the cases when f or when y is greater, and we want their difference to be minimum, so if we would not square the term, then situations in which f

is greater and situation in which y is greater will ancel each other to an extent, and this is not what we want. So, we need to square the term.

Now, for U to be minimum, it must satisfy the following two equations –

= 0 and = 0.

On solving the above two equations, we get two equations, as follows :

?y = nc + m?x, and ?xy = c?x + m?x, which can be rearranged as - m = (n * ?xy - ?x?y) / (n * ?x - (?x)), and c = (?y - m?x) / n,

So, this is how values of m and c for both the cases are obtained, and we can represent a given set of points, by the best possible straight line.

The following code implements the above given algorithm –

## C

`// C Program to find m and c for a straight line given, ` `// x and y ` `#include <stdio.h> ` ` ` `// function to calculate m and c that best fit points ` `// represented by x[] and y[] ` `void` `bestApproximate(` `int` `x[], ` `int` `y[], ` `int` `n) ` `{ ` ` ` `int` `i, j; ` ` ` `float` `m, c, sum_x = 0, sum_y = 0, sum_xy = 0, sum_x2 = 0; ` ` ` `for` `(i = 0; i < n; i++) { ` ` ` `sum_x += x[i]; ` ` ` `sum_y += y[i]; ` ` ` `sum_xy += x[i] * y[i]; ` ` ` `sum_x2 += (x[i] * x[i]); ` ` ` `} ` ` ` ` ` `m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - (sum_x * sum_x)); ` ` ` `c = (sum_y - m * sum_x) / n; ` ` ` ` ` `printf` `(` `"m =% f"` `, m); ` ` ` `printf` `(` `"\nc =% f"` `, c); ` `} ` ` ` `// Driver main function ` `int` `main() ` `{ ` ` ` `int` `x[] = { 1, 2, 3, 4, 5 }; ` ` ` `int` `y[] = { 14, 27, 40, 55, 68 }; ` ` ` `int` `n = ` `sizeof` `(x) / ` `sizeof` `(x[0]); ` ` ` `bestApproximate(x, y, n); ` ` ` `return` `0; ` `} ` |

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## C++

`// C++ Program to find m and c for a straight line given, ` `// x and y ` `#include <cmath> ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// function to calculate m and c that best fit points ` `// represented by x[] and y[] ` `void` `bestApproximate(` `int` `x[], ` `int` `y[], ` `int` `n) ` `{ ` ` ` `float` `m, c, sum_x = 0, sum_y = 0, sum_xy = 0, sum_x2 = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `sum_x += x[i]; ` ` ` `sum_y += y[i]; ` ` ` `sum_xy += x[i] * y[i]; ` ` ` `sum_x2 += ` `pow` `(x[i], 2); ` ` ` `} ` ` ` ` ` `m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - ` `pow` `(sum_x, 2)); ` ` ` `c = (sum_y - m * sum_x) / n; ` ` ` ` ` `cout << ` `"m ="` `<< m; ` ` ` `cout << ` `"\nc ="` `<< c; ` `} ` ` ` `// Driver main function ` `int` `main() ` `{ ` ` ` `int` `x[] = { 1, 2, 3, 4, 5 }; ` ` ` `int` `y[] = { 14, 27, 40, 55, 68 }; ` ` ` `int` `n = ` `sizeof` `(x) / ` `sizeof` `(x[0]); ` ` ` `bestApproximate(x, y, n); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java Program to find m and c for a straight line given, ` `// x and y ` `import` `java.io.*; ` `import` `static` `java.lang.Math.pow; ` ` ` `public` `class` `A { ` ` ` `// function to calculate m and c that best fit points ` ` ` `// represented by x[] and y[] ` ` ` `static` `void` `bestApproximate(` `int` `x[], ` `int` `y[]) ` ` ` `{ ` ` ` `int` `n = x.length; ` ` ` `double` `m, c, sum_x = ` `0` `, sum_y = ` `0` `, ` ` ` `sum_xy = ` `0` `, sum_x2 = ` `0` `; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` `sum_x += x[i]; ` ` ` `sum_y += y[i]; ` ` ` `sum_xy += x[i] * y[i]; ` ` ` `sum_x2 += pow(x[i], ` `2` `); ` ` ` `} ` ` ` ` ` `m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - pow(sum_x, ` `2` `)); ` ` ` `c = (sum_y - m * sum_x) / n; ` ` ` ` ` `System.out.println(` `"m = "` `+ m); ` ` ` `System.out.println(` `"c = "` `+ c); ` ` ` `} ` ` ` ` ` `// Driver main function ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `x[] = { ` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `}; ` ` ` `int` `y[] = { ` `14` `, ` `27` `, ` `40` `, ` `55` `, ` `68` `}; ` ` ` `bestApproximate(x, y); ` ` ` `} ` `} ` |

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## Python3

`# python Program to find m and c for ` `# a straight line given, x and y ` ` ` `# function to calculate m and c that ` `# best fit points represented by x[] ` `# and y[] ` `def` `bestApproximate(x, y, n): ` ` ` ` ` `sum_x ` `=` `0` ` ` `sum_y ` `=` `0` ` ` `sum_xy ` `=` `0` ` ` `sum_x2 ` `=` `0` ` ` ` ` `for` `i ` `in` `range` `(` `0` `, n): ` ` ` `sum_x ` `+` `=` `x[i] ` ` ` `sum_y ` `+` `=` `y[i] ` ` ` `sum_xy ` `+` `=` `x[i] ` `*` `y[i] ` ` ` `sum_x2 ` `+` `=` `pow` `(x[i], ` `2` `) ` ` ` ` ` `m ` `=` `(` `float` `)((n ` `*` `sum_xy ` `-` `sum_x ` `*` `sum_y) ` ` ` `/` `(n ` `*` `sum_x2 ` `-` `pow` `(sum_x, ` `2` `))); ` ` ` ` ` `c ` `=` `(` `float` `)(sum_y ` `-` `m ` `*` `sum_x) ` `/` `n; ` ` ` ` ` `print` `(` `"m = "` `, m); ` ` ` `print` `(` `"c = "` `, c); ` ` ` ` ` `# Driver main function ` `x ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `] ` `y ` `=` `[ ` `14` `, ` `27` `, ` `40` `, ` `55` `, ` `68` `] ` `n ` `=` `len` `(x) ` ` ` `bestApproximate(x, y, n) ` ` ` `# This code is contributed by Sam007. ` |

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## C#

`// C# Program to find m and c for a ` `// straight line given, x and y ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// function to calculate m and c that ` ` ` `// best fit points represented by x[] and y[] ` ` ` `static` `void` `bestApproximate(` `int` `[] x, ` `int` `[] y) ` ` ` `{ ` ` ` `int` `n = x.Length; ` ` ` `double` `m, c, sum_x = 0, sum_y = 0, ` ` ` `sum_xy = 0, sum_x2 = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `sum_x += x[i]; ` ` ` `sum_y += y[i]; ` ` ` `sum_xy += x[i] * y[i]; ` ` ` `sum_x2 += Math.Pow(x[i], 2); ` ` ` `} ` ` ` ` ` `m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - Math.Pow(sum_x, 2)); ` ` ` ` ` `c = (sum_y - m * sum_x) / n; ` ` ` ` ` `Console.WriteLine(` `"m = "` `+ m); ` ` ` `Console.WriteLine(` `"c = "` `+ c); ` ` ` `} ` ` ` ` ` `// Driver main function ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `[] x = { 1, 2, 3, 4, 5 }; ` ` ` `int` `[] y = { 14, 27, 40, 55, 68 }; ` ` ` ` ` `// Function calling ` ` ` `bestApproximate(x, y); ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

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## PHP

`<?php ` `// PHP Program to find m and c ` `// for a straight line given, ` `// x and y ` ` ` `// function to calculate m and ` `// c that best fit points ` `// represented by x[] and y[] ` `function` `bestApproximate(` `$x` `, ` `$y` `, ` `$n` `) ` `{ ` ` ` `$i` `; ` `$j` `; ` ` ` `$m` `; ` `$c` `; ` ` ` `$sum_x` `= 0; ` ` ` `$sum_y` `= 0; ` ` ` `$sum_xy` `= 0; ` ` ` `$sum_x2` `= 0; ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `{ ` ` ` `$sum_x` `+= ` `$x` `[` `$i` `]; ` ` ` `$sum_y` `+= ` `$y` `[` `$i` `]; ` ` ` `$sum_xy` `+= ` `$x` `[` `$i` `] * ` `$y` `[` `$i` `]; ` ` ` `$sum_x2` `+= (` `$x` `[` `$i` `] * ` `$x` `[` `$i` `]); ` ` ` `} ` ` ` ` ` `$m` `= (` `$n` `* ` `$sum_xy` `- ` `$sum_x` `* ` `$sum_y` `) / ` ` ` `(` `$n` `* ` `$sum_x2` `- (` `$sum_x` `* ` `$sum_x` `)); ` ` ` `$c` `= (` `$sum_y` `- ` `$m` `* ` `$sum_x` `) / ` `$n` `; ` ` ` ` ` `echo` `"m ="` `, ` `$m` `; ` ` ` `echo` `"\nc ="` `, ` `$c` `; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$x` `=` `array` `(1, 2, 3, 4, 5); ` ` ` `$y` `=` `array` `(14, 27, 40, 55, 68); ` ` ` `$n` `= sizeof(` `$x` `); ` ` ` `bestApproximate(` `$x` `, ` `$y` `, ` `$n` `); ` ` ` `// This code is contributed by ajit ` `?> ` |

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Output:

m=13.6 c=0.0

**Analysis of above code-**

Auxiliary Space : O(1)

Time Complexity : O(n). We have one loop which iterates n times, and each time it performs constant no. of computations.

**Reference-**

1-Higher Engineering Mathematics by B.S. Grewal.

This article is contributed by **Mrigendra Singh**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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