first off, let's check what's the slope of that line through those two points anyway[tex]\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{7}~,~\stackrel{y_2}{2}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-(-2)}{7-5}\implies \cfrac{2+2}{7-5}\implies \cfrac{4}{2}\implies 2[/tex]now, let's take a peek of what is the slope of that equation then[tex]\bf -5y+kx=6-4x\implies -5y=6-4x-kx\implies -5y=6-x(4+k) \\\\\\ -5y=-x(4+k)+6\implies -5y=-(4+k)x+6\implies y=\cfrac{-(4+k)x+6}{-5}[/tex][tex]\bf y=\cfrac{(4+k)x-6}{5}\implies y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{(4+k)}{5}} x-\cfrac{6}{5}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{since both slopes are the same then}}{\cfrac{4+k}{5}=2\implies 4+k=10}\implies \blacktriangleright k=6 \blacktriangleleft[/tex]