The given line passes through the points (-4,-3) and (4,1) what is the equation in point-slope form, of the line that is perpendicular to the given line and passes through the point (-4,3)?

let's bear in mind that perpendicular lines have negative reciprocal slopes...... hmmmm what's the slope of the given line anyway?[tex]\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-3)}{4-(-4)}\implies \cfrac{1+3}{4+4}\implies \cfrac{4}{8}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill[/tex][tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{1}{2}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{2}{1}}\qquad \stackrel{negative~reciprocal}{-\cfrac{2}{1}\implies -2}}[/tex]so, we're really looking for a line whose slope is -2 and runs through (-4 , 3)[tex]\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})~\hspace{10em} slope = m\implies -2 \\\\\\ \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-(-3)=-2[x-(-4)] \implies y+3=-2(x+4) \\\\\\ y+3=-2x-8\implies y=-2x-11[/tex]