The equation is $(y-1)=m(x-0)$
$\Rightarrow m x-y+1=0$
$\Rightarrow \frac{1}{\sqrt{m^{2}+1}}=\frac{3}{5}$
$\left[\because d=\left|\frac{x_{1} a+y_{1} b+c}{\sqrt{a^{2}+b^{2}}}\right|\right.$ at point $\left.\left(x_{1}, y_{1}\right)\right]$
$\Rightarrow m^{2}+1=\frac{25}{9} \Rightarrow m^{2}=\frac{16}{9} \Rightarrow m=\pm \frac{4}{3}$
Thus, the required equation of line is Then,
$-\frac{4}{3} x-y+1=0$ [from Eq.(i) ]
$\Rightarrow 4 x+3 y-3=0 \Rightarrow 4 x+3 y=3$