Dose finding slope/intercept using the formula of m,b gives best fit line always In linear regression? In liner regression We have to fit different lines and chose one with minimum error so What is the motive of having a formula for m,b that can give slope and intercept value in the regression line ,when it cannot give best fit line directly ?

1.Consider i applied the value in dataset on the  formula of m,b and found the regression line  yhat = 17.5835x+6 and for example just assume   error calculated for this line was  3
2.Consider i fit another line randomly (i am not using the formula of m,b to find value of m,b  assume m,b value for this random line was 16,3) my 2nd regression line is yhat = 16x+3and  for example just assume   error calculated for this line was 1.5
Linear Regression Goal : to choose best fit line that has minimum error
so my second line is better than the 1st line in this case
What is the point of having a formula which gives value for slope &quot;m&quot;, intercept &quot;b&quot; when it cannot give best fit line directly ?
OR is my understanding incoorect Dose finding slope/intercept using the formula of m,b  gives best line always ?
if its YES   then there is no need to try mulitple lines and calculate error  and choose line with min error
if its No  then whats the point of having a formula for slope m,intercept b when it cannot give the best fit line . dose that mean maths/stats community need to change this forumla for slope,intercept
The formulae you mentioned gives the coefficients of the line of best fit.The values are derived using the least squares method, where the goal is to minimize the sum of squared errors. Following is the derivation for the values of m and b.
Let the line of best fit be $$\hat{y} = m*x + b$$ We then  try to find the coefficients m and b which minimize the sum of squared errors between the actual value y and the observed value $\hat{y}$. \begin{align}
SSE &amp;= \sum_{i=1}^{n}(y_{i}-\hat{y_{i}})^2 \\
&amp;=\sum_{i=1}^{n}(y_{i}-m*x_{i}-b)^2
\end{align}
Taking the first derivative of SSE with respect to c and equating to zero.
\begin{align}
\frac{\partial SSE}{\partial b} &amp;= \sum_{i=1}^{n}-2*(y_{i}-m*x_{i}-b)\\
0 &amp;= \sum_{i=1}^{n}-2*(y_{i}-m*x_{i}-b)
\end{align}
Therefore we get c as $$ b = \bar{y} - m*\bar{x}$$ Similarly in order to find m we take the partial derivative of SSE with respect to m and equate it to zero.
\begin{align}
\frac{\partial SSE}{\partial m} &amp;= \sum_{i=1}^{n}-2x_{i}*(y_{i}-m*x_{i}-b)\\
0 &amp;= \sum_{i=1}^{n}-2x_{i}*(y_{i}-m*x_{i}-b)\\
0 &amp;= \sum_{i=1}^{n}x_{i}*(y_{i}-m*x_{i}-b)\\
0 &amp;= \sum_{i=1}^{n}x_{i}*y_{i} - \sum_{i=1}^{n}m*x_{i}^2 - \sum_{i=1}^{n}b*x_{i}
\end{align}
Substituting b and solving for m we get $$m = \frac{n\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2}$$