The distance between the lines $3x - 2y = 1$ and $6x - 4y + 9 = 0$ is

The vertices of a triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to the origin is:

If the line $x-y+1=0$ cuts the lines $2x+2y+3=0$ and $3x+3y+2=0$ at the points $A$ and $B$ respectively,then $AB=$

If the points $(1, 2)$ and $(3, 4)$ lie on the same side of the straight line $3x - 5y + a = 0$,then $a$ lies in the set

If a line $l$ passes through $(k, 2k), (3k, 3k)$ and $(3, 1)$,where $k \neq 0$,then the distance from the origin to the line $l$ is

If $(\sin \theta, \cos \theta)$ and $(3, 2)$ lie on the same side of the line $x + y = 1$,then $\theta$ lies in the interval: