What is the angle in degrees between the lines $x = 9$ and $x - \sqrt{3}y + 7 = 0$?

$A$ straight line $L$ through the point $(3, -2)$ is inclined at an angle of $60^o$ to the line $\sqrt{3}x + y = 1$. If $L$ also intersects the $x$-axis,then the equation of $L$ is

The acute angle between the lines given by $y - \sqrt{3}x + 1 = 0$ and $\sqrt{3}y - x + 7 = 0$ is (in $^{\circ}$)

The lines $p(p^2 + 1)x - y + q = 0$ and $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$ are perpendicular to a common line for :

If the lines $2x + 3ay - 1 = 0$ and $3x + 4y + 1 = 0$ are mutually perpendicular,then the value of $a$ will be

The line passing through the points $(3, -4)$ and $(-2, 6)$ and the line passing through $(-3, 6)$ and $(9, -18)$ are: