Answer
$m\Sigma_{i=1}^{n}x_i+bn=\Sigma_{i=1}^{n} y_i$
and
$m\Sigma_{i=1}^{n}x_i^2+b\Sigma_{i=1}^{n}x_i=\Sigma_{i=1}^{n}x_i y_i$
Work Step by Step
Given: $d_i=y_i-(mx_i+b)$
These are the vertical deviation of the point $(x_i,y_i)$ from the straight line.
Now,
$\dfrac{\partial L}{\partial m}=\Sigma_{i=1}^{n}\dfrac{\partial }{\partial m}[y_i-(mx_i+b)]$
or, $=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-x_i)$
or, $\dfrac{\partial L}{\partial m}=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-x_i)$
Thus,
$m\Sigma_{i=1}^{n}x_i^2+b\Sigma_{i=1}^{n}x_i=\Sigma_{i=1}^{n}x_i y_i$
Also,
Given: $d_i=y_i-(mx_i+b)$
$\dfrac{\partial L}{\partial b}=\Sigma_{i=1}^{n}\dfrac{\partial }{\partial m}[y_i-(mx_i+b)]$
or, $=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-1)$
or, $\dfrac{\partial L}{\partial b}=\Sigma_{i=1}^{n}2[y_i-(mx_i+b)](-1)$
Thus,
$m\Sigma_{i=1}^{n}x_i+bn=\Sigma_{i=1}^{n} y_i$
