A give line has the equation 10x+2y=-2. What is the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12)? Y=(________)x + 12​

first off, let's solve the "given" line for "y", that way we can put it in slope-intercept form to see what its slope might be.[tex]\bf 10x+2y=-2\implies 2y=-10x-2\implies y=\cfrac{-10x-2}{2} \\\\\\ y=\cfrac{-10x}{2}-\cfrac{2}{2}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-5}x-1\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}[/tex]well, a parallel line to that one, will have the same exact slope, so then, we're really looking for the equation of a line whose slope is -5 and runs through (0 , 12)[tex]\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{12})~\hspace{10em} slope = m\implies -5 \\\\\\ \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-12=-5(x-0) \\\\\\ y-12=-5x\implies y=-5x+12[/tex]