Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$. (in $/5$)

$A$ straight line through the origin $O$ meets the parallel lines $4x + 2y = 9$ and $2x + y + 6 = 0$ at $P$ and $Q$ respectively. The point $O$ divides the segment $PQ$ in the ratio

The length of the perpendicular from the point $(3, 1)$ to the line $4x + 3y + 20 = 0$ is:

The base of an equilateral triangle is along the line given by $3x + 4y = 9$. If a vertex of the triangle is $(1, 2)$,then the length of a side of the triangle is

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

If the point $(1, a)$ lies between the lines $x + y = 1$ and $2(x + y) = 3$,then in which interval does $a$ lie?