check the picture below.
so red line of BD is perpendicular to AC, hmmmm let's firstly find the slope of AC, bearing in mind that perpendicular lines have negative reciprocal slopes.
[tex]\bf A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{18}~,~\stackrel{y_2}{-8}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-8-(-2)}{18-(-4)}\implies \cfrac{-8+2}{18+4} \\\\\\ \cfrac{-6}{22}\implies -\cfrac{3}{11} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{11}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{11}{3}}\qquad \stackrel{negative~reciprocal}{\cfrac{11}{3}}}\impliedby \textit{BD's slope}[/tex]
so we're really looking for the equation of a line whose slope is 11/3 and runs through B(4,4). 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 B(\stackrel{x_1}{4}~,~\stackrel{y_1}{4})~\hspace{10em} slope = m\implies \cfrac{11}{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{11}{3}(x-4)[/tex]
[tex]\bf \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}\textit{ to do away with the denominators}}{3(y-4)=3\left( \cfrac{11}{3}(x-4) \right)\implies 3y-12=11(x-4)} \\\\\\ 3y-12=11x-44\implies 3y=11x-32 \\\\\\ -11x+3y=-32\implies \blacktriangleright 11x - 3y= 32\blacktriangleleft[/tex]
