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

Let the lines be $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$. For the point $A(3, -5)$,which of the following is true?

The line $L$ is given by $\frac{x}{5} + \frac{y}{b} = 1$ and passes through the point $(13, 32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$. Then the distance between $L$ and $K$ is

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the lines $x \cos \theta - y \sin \theta = k \cos 2 \theta$ and $x \sec \theta + y \csc \theta = k$ respectively,prove that $p^{2} + 4q^{2} = k^{2}$.

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

If $O$ is the origin and $P, Q$ are points on the line $3x + 4y + 15 = 0$ such that $OP = OQ = 9$, then the area of $\triangle OPQ$ is (in $\sqrt{2}$)