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The distance between the parallel lines y = 2x + 4 and 6x = 3y + 5 is : 
1. 1 unit
2. \(\frac{17\sqrt 5}{15}\) unit
3. \(\frac{3}{\sqrt 5}\) unit
4. \(\frac{17}{\sqrt 5}\) unit

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Correct Answer - Option 2 : \(\frac{17\sqrt 5}{15}\) unit

Concept:

The distance between the two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is given by:

\(d=\frac{\left | c_1-c_2 \right |}{\sqrt{a^2+b^2}}\)

Calculation:

The given equations are y = 2x + 4 and 6x = 3y + 5.

These can also be written as 2x - y + 4 = 0 and 2x - y - 5/3 = 0.

Now, since the lines are parallel.

\(\Rightarrow d=\frac{\left | c_1-c_2 \right |}{\sqrt{a^2+b^2}}\)

\(\Rightarrow d=\frac{\left | 4+\frac{5}{3} \right |}{\sqrt{2^2+3^2}}\)

\(\Rightarrow d=\frac{\left | \frac{17}{3} \right |}{\sqrt{5}}\)

\(\Rightarrow d=\frac{17\sqrt{5}}{15}\)

Hence, the distance between the parallel lines is ​​\(\frac{17\sqrt 5}{15}\) unit.

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