The distance between the parallel lines $3x + 4y + 7 = 0$ and $3x + 4y - 5 = 0$ is

Find the length of the perpendicular drawn from the point $(2, 1)$ to the line $3x - 4y + 8 = 0$.

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is

Find the length of the perpendicular drawn from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$.

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the straight lines $x \sec \alpha + y \operatorname{cosec} \alpha = 10$ and $x \cos \alpha - y \sin \alpha = 10 \cos 2 \alpha$,then $4 p^2 + q^2 =$

Let $\alpha$ be the distance between the lines $-x + y = 2$ and $x - y = 2$,and $\beta$ be the distance between the lines $4x - 3y = 5$ and $6y - 8x = 1$,then: