check the picture below
so.. .if we use those 2 points, from the "best fit line"
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 10}}\quad ,&{{ 20}})\quad
% (c,d)
&({{ 20}}\quad ,&{{ 40}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{40-20}{20-10}\implies \cfrac{20}{10}\implies 2[/tex]
[tex]\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad
\begin{array}{llll}
\begin{cases}
y_1=20\\
x_1=10\\
m=2
\end{cases}\\
\end{array}\implies y-20=2(x-10)\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-20=2x-20\implies y=2x-20+20\implies \boxed{y = 2x}[/tex]
