[tex]\bf 3y=-4x+2\implies y=\cfrac{-4x+2}{3} \\\\\\ y=\stackrel{\stackrel{slope}{\downarrow }}{-\cfrac{4}{3}}x+\cfrac{2}{3}\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]
so the slope of that line above is really -4/3, now
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{4}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{4}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{4}\implies \blacktriangleright \cfrac{3}{4} \blacktriangleleft}}[/tex]
Answer:
[tex]\frac{3}{4}[/tex]
Step-by-step explanation:
Verified correct with test results.
