The coordinates of the point on the line $x+y+3=0$,whose distance from the line $x+2y+2=0$ is $\sqrt{5}$ units,are

If $p$ and $q$ are the perpendicular distances from the origin to the straight lines $x \sec \theta - y \operatorname{cosec} \theta = a$ and $x \cos \theta + y \sin \theta = a \cos 2 \theta$,then

Given the lines $y + 2x = 3$ and $y + 2x = 5$ cut the axes at $A, B$ and $C, D$ respectively.
Statement-$1$ : $ABDC$ forms a quadrilateral and the point $(2, 3)$ lies inside the quadrilateral.
Statement-$2$ : The point $(2, 3)$ lies between the two parallel lines.

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:

The distance between the lines given by $3x + 4y = 9$ and $6x + 8y = 15$ is (in $units$)

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