Answer :
[tex]k:\ y = x-11\ \ \ \Leftrightarrow\ \ \ x-y-11=0\\ and\\ l:\ y = x-7\ \ \ \Leftrightarrow\ \ \ x-y-7=0\\\\the\ distance:\\\\ d(k;l)= \frac{\big{|-11-(-7)|}}{\big{ \sqrt{1^2+1^2} }} =\frac{\big{|-11+7|}}{\big{ \sqrt{2} }} =\frac{\big{|-4|}}{\big{ \sqrt{2} }} =\frac{\big{4\cdot \sqrt{2} }}{\big{ \sqrt{2}\cdot \sqrt{2} }} =\frac{\big{4 \sqrt{2} }}{\big{2 }} =2 \sqrt{2} [/tex]
[tex]Given \ the \ equations \ of \ two \ non-vertical \ parallel \ lines:\\\\y = mx+b_1\\y = mx+b_2\\\\the \ distance \ between \ them \ can \ be \ expressed \ as : \\\\d= \frac{|b_{1}-b_{2}|}{ \sqrt{ m^2+1} }[/tex]
[tex]y = x-11 \\ y = x-7 \\\\\\d= \frac{| -11- (-7)|}{ \sqrt{ 1^2+1} } =\frac{| -11+7|}{ \sqrt{ 1+1} } = \frac{|-4|}{ \sqrt{2} } = \frac{4}{ \sqrt{2} }\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}[/tex]
[tex]y = x-11 \\ y = x-7 \\\\\\d= \frac{| -11- (-7)|}{ \sqrt{ 1^2+1} } =\frac{| -11+7|}{ \sqrt{ 1+1} } = \frac{|-4|}{ \sqrt{2} } = \frac{4}{ \sqrt{2} }\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}[/tex]
