A) 1
B) \[3/\sqrt{5}\]
C) \[\frac{17\sqrt{5}}{15}\]
D) \[\frac{17}{\sqrt{3}}\]
Correct Answer: C
Solution :
Since lines \[y-2x-4=0\]and \[y-2x+\frac{5}{3}=0\] are parallel. Distance \[=\left| \frac{{{c}_{2}}-{{c}_{1}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|=\left| \frac{\frac{5}{3}+4}{\sqrt{4}+1} \right|=\frac{17\sqrt{5}}{15}\]
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