What is the distance between the lines 3x - 4y - 5 = 0 and 6x - 8y + 10 = 0?

The lines given are,

3x - 4y - 5 = 0   ----(1)

6x - 8y + 10 = 0  -----(2)

(1) * 2, we get,

6x - 8y - 10 = 0 ----(1)

We can see that (1) and (2) are parallel to each other as the slopes of the lines are equal.

The distance can be calculated using the formula,

$$dis\tan ce\ =\ \dfrac{\left|c_1\ -\ c_2\right|}{\sqrt{\ a^2\ +\ b^2}}$$

$$dis\tan ce\ =\ \dfrac{\left|10\ -\left(-10\right)\right|}{\sqrt{\ 6^2\ +\ \left(-8\right)^2}}$$

$$dis\tan ce\ =\ \dfrac{20}{\sqrt{\ 36\ +\ 64}}\ =\ \dfrac{20}{10}\ =\ 2$$

The correct answer is option B.