The perpendicular form of the line  $3x+4y-5=0$ is

$x\cos \alpha +y\sin \alpha =1\text{\hspace{0.17em}\hspace{0.17em}where\hspace{0.17em}\hspace{0.17em}}\cos \alpha =\frac{3}{5},\sin \alpha =\frac{4}{5}$

$x\cos \alpha -y\sin \alpha =1\text{\hspace{0.17em}\hspace{0.17em}where\hspace{0.17em}\hspace{0.17em}}\cos \alpha =\frac{3}{5},\sin \alpha =\frac{4}{5}$

$x\cos \alpha +y\sin \alpha =1\text{\hspace{0.17em}\hspace{0.17em}where\hspace{0.17em}\hspace{0.17em}}\cos \alpha =\frac{4}{5},\sin \alpha =\frac{3}{5}$

$x\cos \alpha -y\sin \alpha =1\text{\hspace{0.17em}\hspace{0.17em}where\hspace{0.17em}\hspace{0.17em}}\cos \alpha =\frac{4}{5},\sin \alpha =\frac{3}{5}$