The given parallel lines are $l(x+y)+p=0\: and \: l(x+y)-r=0$

Here $A=l, B = l, C_1=p \:and \: C_2=r$.

Hence substituting the values in $d = \large\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$ we get,

$ d = \large\frac{|p-(-r)|}{\sqrt{l^2+l^2}}$$ = \large\frac{|p+r|}{l\sqrt 2}$

Hence the required distance is $ \large\frac{1}{\sqrt 2}$$ \bigg| \large\frac{p+r}{l} \bigg|$ units.