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 $\sqrt{2}$)

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 $3x + 4y = 9$ and $6x + 8y = 15$ is

$A$ straight line is parallel to the lines $3x - y - 3 = 0$ and $3x - y + 5 = 0$ and lies between them. If its distance from these lines is in the ratio $3 : 5$,then its equation 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 $(\omega)$.

The length of the perpendicular from $(1, -2)$ to the line $12x + 5y + 63 = 0$ is