If the distance between the lines $y=mx+1$ and $y=mx+6$ is $\sqrt{5}$  units, then the value of $‘\text{m}’$  is

$c_1=1,c_2=6,a=m,b=-1$

Distance $=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$

$=\left|\frac{1-6}{\sqrt{m^2+{\left(-1\right)}^2}}\right|$

$=\left|\frac{-5}{\sqrt{m^2+1}}\right|=\sqrt{5}$

$=\frac{5}{\sqrt{m^2+1}}=\sqrt{5}$

$5=\sqrt{5\left(m^2+1\right)}$

$25=5\left(m^2+1\right)$

$\frac{25}{5}=m^2+1$

$5=m^2+1$

$m^2=4$

$m=\pm 2$