bearing in mind that parallel lines have the same exact slope, hmmm what is the slope of y = -10x + 3 anyway?
[tex]\bf y=\stackrel{\downarrow }{-10}x+3\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] ~\dotfill[/tex]
[tex]\bf \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 (\stackrel{x}{5},\stackrel{y}{-6})~\hfill \begin{array}{llll} y-(-6)=-10(x-5) \\\\\\ y+6=-10(x-5) \end{array}[/tex]
