Help! Use the given conditions to write an equation for the line in point-slope form.Passing through (-5, -7) and (-8, -6)A) y− 7 = −13(x−5) or y −6 = −13(x−8)B) y+7= −13(x+8) or y+6 = −13(x+5)C) y+7 = −13(x+5) or y+6 = −13(x+8)D) y+7 = −13(x+5) or y +6 = −13(x+7)

[tex]\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{-6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-6-(-7)}{-8-(-5)}\implies \cfrac{-6+7}{-8+5}\implies \cfrac{1}{-3}\implies -\cfrac{1}{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-(-7)=-\cfrac{1}{3}[x-(-5)]\implies y+7=-\cfrac{1}{3}(x+5)[/tex][tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_2=m(x-x_2) \\\\ \cline{1-1} \end{array}\implies y-(-6)=-\cfrac{1}{3}[x-(-8)]\implies y+6=-\cfrac{1}{3}(x+8)[/tex]