Answer: [tex]x+7y+5(1+\sqrt{2})\ and\ x+7y+5(1-\sqrt{2})[/tex]
Explanation:
General equation of line parallel to the given line x+7y+2=0 is
x+7y+c=0
This line must be unit distance from point (2,-1)
Point (2,-1) may be either side of line so we can write
[tex]\frac{\left | 2+7(-1)+c \right |}{\sqrt{1^2+7^2}}=1\\\frac{\left | c-5 \right |}{\sqrt{50}}=1\\\left | c-5 \right |=\sqrt{50}\\c-5=\pm\sqrt{50}=\pm5\sqrt{2}\\c=5\pm5\sqrt{2}[/tex]
So, there can be 2 line i.e.
[tex]x+7y+5(1+\sqrt{2})\ and\ x+7y+5(1-\sqrt{2})[/tex]
