If the medians AD and BE of the triangle with vertices $A(0, b), B (0, 0)$ and $C (a, 0)$are perpendicular,
then an equation of the line through (a, b) perpendicular to AC is

Slope of AD $=\frac{b-0}{0-a/2}=-\frac{2b}{a}$

Slope of $BE=\frac{b/2}{a/2}=\frac{b}{a}$

Since AD and BE are perpendicular

$$\begin{gathered} \frac{-2b}{a}\cdot \frac{b}{a}=-1 \\ \Rightarrow a^2=2b^2 \left(i\right) \end{gathered}$$  

Slope of AC is – b/a 

$\therefore$ Equation of the line through (a, b) perpendicular to AC is

$\begin{array}{l}y-b=\frac{c}{b}(x-a)\\ \Rightarrow ax-by=a^2-b^2\end{array}$

$\Rightarrow \pm \sqrt{2}bx-by=b^2$ from (i)

$\pm \sqrt{2}x-y=b\Rightarrow y=\pm \sqrt{2}x-b$

So an equation of required line is $y=\sqrt{2}x-b$