If the point $(3, -4)$ divides the segment of the line between the $X$-axis and $Y$-axis in the ratio $2 : 3$,then the equation of the line is:

$A$ line $L$ passes through the point $P(1, 2)$ and makes an angle of $60^{\circ}$ with the positive $X$-axis. $A$ and $B$ are two points lying on $L$ at a distance of $4$ units from $P$. If $O$ is the origin,then the area of $\triangle OAB$ is

$A$ line passing through the point $P(a, 0)$ makes an acute angle $\alpha$ with the positive $x$-axis. Let this line be rotated about the point $P$ through an angle $\frac{\alpha}{2}$ in the clockwise direction. If in the new position,the slope of the line is $2-\sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$,then the value of $3a^2 \tan^2 \alpha - 2\sqrt{3}$ is

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 = $

Let a line intersect the co-ordinate axes in points $A$ and $B$ such that the area of the triangle $OAB$ is $12$ sq. units. If the line passes through the point $(2,3)$,then the equation of the line is

$A$ straight line parallel to the line $y=\sqrt{3}x$ passes through $Q(2,3)$ and cuts the line $2x+4y-27=0$ at $P$. Then the length of the line segment $PQ$ is