Answer :
When the equations of two parallel lines are given in slope-intercept form, as:
[tex]\begin{gathered} y=mx+b_1 \\ y=mx+b_2 \end{gathered}[/tex]The distance between those two lines is given by the formula:
[tex]d=\frac{|b_1-b_2|}{\sqrt[]{1+m^2}}[/tex]For the given equations:
[tex]\begin{gathered} y=\frac{1}{3}x+2 \\ y=\frac{1}{3}x-8 \end{gathered}[/tex]We can see that m=1/3, b₁=2 and b₂=-8. Then, the distance between those two lines, is:
[tex]\begin{gathered} d=\frac{|2-(-8)|}{\sqrt[]{1+(\frac{1}{3})^2}} \\ =\frac{|2+8|}{\sqrt[]{1+\frac{1}{9}}} \\ =\frac{10}{\sqrt[]{\frac{10}{9}}} \\ =10\cdot\sqrt[]{\frac{9}{10}} \\ =\frac{\sqrt[]{10^2}}{\sqrt[]{10}}\cdot\sqrt[]{9} \\ =\sqrt[]{\frac{10^2}{10}}\times3 \\ =\sqrt[]{10}\times3 \\ =3\cdot\sqrt[]{10} \\ \approx9.487\ldots \end{gathered}[/tex]Therefore, the distance between those two lines is:
[tex]3\cdot\sqrt[]{10}[/tex]