What is the y-intercept of the line perpendicular to the line y = 4/3x + 1 that includes the point (4, 1)?

[tex]\bf y=\stackrel{\downarrow }{\cfrac{4}{3}}x+1\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 \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}}}[/tex]so then, we know this line has a slope of -3/4 and runs through (4 , 1)[tex]\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{1})~\hspace{10em} slope = m\implies -\cfrac{3}{4} \\\\\\ \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-1=-\cfrac{3}{4}(x-4)\implies y-1=-\cfrac{3}{4}x+3[/tex][tex]\bf y=-\cfrac{3}{4}x\stackrel{\downarrow }{+4}\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}~\hfill \stackrel{\textit{y-intercept}}{(0,4)}[/tex]