Question 1134741
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You have received two good solutions to your problem that use similar basic algebraic techniques, utilizing the fact that the slopes of perpendicular lines (except horizontal and vertical lines) have slopes that are negative reciprocals.<br>
There is a shortcut to solving this kind of problem.<br>
(If the equation doesn't have to be in slope-intercept form, this shortcut gets you to AN equation of the line immediately; in this problem, since an equation in slope-intercept form is required, the shortcut doesn't save much time.)<br>
Given a line with equation Ax+By=C, any line parallel to that given line has an equation of the form Ax+By=D (where D is some probably different constant); and any line PERPENDICULAR to that given line has an equation of the form Bx-Ay=E (where again E is some constant).<br>
Note the coefficients are switched; and one of them changes sign.<br>
In your example, the given line has the equation x+2y=-4.  Any line perpendicular to that line has an equation of the form 2x-y=C.<br>
Plug in the coordinates of the given point the find C; then convert the equation to slope-intercept form.<br>
{{{2(-3)-(-5) = -6+5 = -1}}}, so the equation is<br>
{{{2x-y = -1}}}
{{{y = 2x+1}}}