keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm so, what is the slope of y = 1/3x + 2 anyway? well
[tex]\bf y=\stackrel{slope}{\cfrac{1}{3}}x+2\qquad therefore
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\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{1}}\implies -3}[/tex]
so, we're really looking for the equation of a line whose slope is -3 and runs through 6, -3.
[tex]\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{-3})\qquad \qquad \qquad
slope = m\implies -3
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% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-3)=-3(x-6)
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y+3=-3x+18\implies y=-3x+15[/tex]
