well, we know the slope is 3/2, what's the midpoint of those anyway?
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 6}}\quad ,&{{ 3}})\quad
% (c,d)
&({{ -2}}\quad ,&{{ 11}})
\end{array}\qquad
% coordinates of midpoint
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{-2+6}{2}~~,~~\cfrac{11+3}{2} \right)\implies (2,7)[/tex]
so, what's the equation of a line whose slope is 3/2 and runs through 2,7?
[tex]\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ 7}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{3}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-7=\cfrac{3}{2}(x-2)
\\\\\\
y-7=\cfrac{3}{2}x-3\implies y=\cfrac{3}{2}x+4[/tex]
