let's first off find the slope of AB
[tex]\bf A(\stackrel{x_1}{4}~,~\stackrel{y_1}{7})\qquad B(\stackrel{x_2}{-6}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-7}{-6-4}\implies \cfrac{-6}{-10}\implies \cfrac{3}{5}[/tex]
now let's find the midpoint of AB
[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{4}~,~\stackrel{y_1}{7})\qquad B(\stackrel{x_2}{-6}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-6+4}{2}~,~\cfrac{1+7}{2} \right)\implies \left( \cfrac{-2}{2}~,~\cfrac{8}{2} \right)\implies (-1,4)[/tex]
now, a line perpendicular to AB will
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{3}{5}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{5}{3}}\qquad \stackrel{negative~reciprocal}{-\cfrac{5}{3}}}[/tex]
so, we're really looking for a line whose slope is -5/3 and runs through (-1, 4)
[tex]\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{4})~\hspace{10em} slope = m\implies -\cfrac{5}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-\cfrac{5}{3}[x-(-1)] \implies y-4=-\cfrac{5}{3}(x+1) \\\\\\ y-4=-\cfrac{5}{3}x-\cfrac{5}{3}\implies y=-\cfrac{5}{3}x-\cfrac{5}{3}+4\implies y=-\cfrac{5}{3}x+\cfrac{7}{3}[/tex]
now let's convert that to standard form, keeping in mind that
standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
[tex]\bf \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y)=3\left( -\cfrac{5}{3}x+\cfrac{7}{3} \right)}\implies 3y=-5x+7\implies \blacktriangleright 5x+3y=7 \blacktriangleleft[/tex]
