Question 828236
The given line is {{{y=(1/3)x-2/3}}} so any general point of the line is (x,(1/3)(x-2)).


You want another line, {{{y=(1/3)x+b}}}, to be distance {{{2*sqrt(10)}}} from the given line.  General point of the wanted line is (x,(1/3)x+b).


Distance Formula:  {{{sqrt((x-x)^2+((1/3)(x-2)-((1/3)x+b))^2)=2*sqrt(10)}}}
and what is needed is a formula for b in terms of x or no terms of x.
'
{{{(0+((1/3)(x-2)-((1/3)x+b))^2)=4*10}}}
{{{(x/3-2/3-x/3-b)^2=40}}}
{{{(-2/3-b)^2=40}}}
{{{(b+2/3)^2=40}}}
{{{b+2/3=0+- 2*sqrt(10)}}}
{{{b=-(2/3)+- 2*sqrt(10)}}}


Two possible equations then would be {{{highlight(y=(1/3)x-(2/3)-2*sqrt(10))}}} and {{{highlight(y=(1/3)x-2/3+2*sqrt(10))}}}