We want to find the equation of the normal line of [tex]y=\dfrac{20-x}{3x}[/tex] at the point [tex]P=(x_0,y_0)[/tex], where [tex]y_0=3[/tex]. First calculate [tex]x_0[/tex]. We have:
[tex]y_0=f(x_0)=\dfrac{20-x_0}{3x_0}\\\\\\3=\dfrac{20-x_0}{3x_0}\quad|\cdot3x_0\\\\\\3\cdot3x_0=20-x_0\\\\9x_0=20-x_0\\\\9x_0+x_0=20\\\\10x_0=20\quad|:2\\\\\boxed{x_0=2}[/tex]
Now, when we know that [tex]P=(x_0,y_0)=(2,3)[/tex] we can write an equation of the normal line as:
[tex]\boxed{y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)}[/tex]
Calculate [tex]f'(x_0)[/tex]:
[tex]f'(x)=\left(\dfrac{20-x}{3x}\right)'=\left(\dfrac{20}{3x}\right)'-\left(\dfrac{x}{3x}\right)'=\dfrac{20}{3}\cdot\left(\dfrac{1}{x}\right)'-\left(\dfrac{1}{3}\right)'=\\\\\\=\dfrac{20}{3}\cdot\left(-\dfrac{1}{x^2}\right)-0=\boxed{-\dfrac{20}{3x^2}}\\\\\\\\f'(x_0)=f'(2)=-\dfrac{20}{3\cdot2^2}=-\dfrac{20}{3\cdot4}=-\dfrac{20}{12}=\boxed{-\frac{5}{3}}[/tex]
and the equation of the normal line:
[tex]y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\y-3=-\dfrac{1}{-\frac{5}{3}}(x-2)\\\\\\y-3=\dfrac{3}{5}(x-2)\\\\\\y-3=\dfrac{3}{5}x-\dfrac{6}{5}\\\\\\y=\dfrac{3}{5}x-\dfrac{6}{5}+3\\\\\\\boxed{y=\dfrac{3}{5}x+\dfrac{9}{5}}[/tex]
