If  the line $4x+3y+1=0$  cuts the axes at $A$  and $B$ ,  then the equation of perpendicular bisector of $AB$  is

$A=\left(\frac{-1}{4},0\right),B=\left(0,\frac{-1}{3}\right)$

Slope $\left(m\right)=\frac{-a}{b}=\frac{-4}{3}$

Slope of a perpendicular line $\left(m_1\right)=\frac{-1}{m}$

$=\frac{-1}{\left(\frac{-4}{3}\right)}$ $=\frac{-3}{-4}=\frac{3}{4}$

Point $C$ $=$  midpoint of $AB$

$=\left[\frac{\frac{-1}{4}+0}{2},\frac{0-\frac{1}{3}}{2}\right]$ $=\left[\frac{-1}{8},\frac{-1}{6}\right]$

The line perpendicular bisector of $4x+3y+1=0$  is

$y-y_1=m_1\left(x-x_1\right)$

$\left(y+\frac{1}{6}\right)=\frac{3}{4}\left(x+\frac{1}{8}\right)$

$\frac{6y+1}{6}=\frac{3}{4}\left(\frac{8x+1}{8}\right)$

$\frac{6y+1}{3}=\frac{3\left(8x+1\right)}{16}$

$96y+16=72x+9$

$72x-96y-7=0$