Find the distance between the lines $3x + 2y + 7 = 0$ and $6x + 4y + 3 = 0$.

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:

Let the origin be the centroid of an equilateral triangle $ABC$ and one of its sides be along the straight line $x+y=3$. If $R$ and $r$ are its circumradius and inradius respectively,then $R+r=$

The coordinates of the two points lying on $x+y=4$ and at a unit distance from the straight line $4x+3y=10$ are

$A$ line $L_1$ passing through the point of intersection of the lines $x-2y+3=0$ and $2x-y=0$ is parallel to the line $L_2$. If $L_2$ passes through the origin and also through the point of intersection of the lines $3x-y+2=0$ and $x-3y-2=0$,then the distance between the lines $L_1$ and $L_2$ is

Reduce the equation $x-y=4$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.