If $x \cos \alpha + y \sin \alpha = p$ is the normal form of the equation of a straight line $x + \sqrt{3} y + 4 = 0$ and $a, b$ are respectively $X$ and $Y$ intercepts of this line,then $\sqrt{3} \pi b p - 3 a \alpha = $

Find the equations of the lines parallel to the axes and passing through the point $(-2, 3)$.

The polar equation of the line perpendicular to the line $\sin \theta - \cos \theta = \frac{1}{r}$ and passing through the point $\left(2, \frac{\pi}{6}\right)$ is

The equation of the line perpendicular to the line $\frac{x}{a} - \frac{y}{b} = 1$ and passing through the point at which it cuts the $x$-axis,is

In what direction should a line be drawn through the point $(1, 2)$ so that its point of intersection with the line $x + y = 4$ is at a distance of $\frac{\sqrt{6}}{3}$ from the given point (in $^\circ$)?

$A$ line through $A(-5, -4)$ meets the lines $x + 3y + 2 = 0$,$2x + y + 4 = 0$,and $x - y - 5 = 0$ at $B$,$C$,and $D$ respectively. If $\left( \frac{15}{AB} \right)^2 + \left( \frac{10}{AC} \right)^2 = \left( \frac{6}{AD} \right)^2$,then the equation of the line is