The number of points having a distance of $\sqrt{5}$ from the straight line $x-2y+1=0$ and a distance of $\sqrt{13}$ from the line $2x+3y-1=0$ is:

$A$ straight line through the origin $O(0, 0)$ meets the lines $4x + 3y - 10 = 0$ and $8x + 6y + 5 = 0$ at points $A$ and $B$ respectively. Then $O$ divides the segment $AB$ in the ratio:

$A$ straight line $L \equiv 0$ passing through the point $A=(-5,-4)$ and having slope $\tan \theta$ meets the lines $x+3y+2=0$ and $2x+y+4=0$ respectively at the points $B$ and $C$. If $\frac{100}{AC^2}-\frac{225}{AB^2}=4 \cos 2\theta+\sin 2\theta$,then the slope of the line $L \equiv 0$ is

$A$ is a point on either of two lines $y + \sqrt{3} |x| = 2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection. The coordinates of the foot of the perpendicular from $A$ on the bisector of the angle between them are

$A$ line is such that its segment between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$,then its equation is

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the lines $x \sec \alpha + y \csc \alpha = 2a$ and $x \cos \alpha - y \sin \alpha = a \cos 2\alpha$ respectively,then what is the value of $\left( \frac{p_1}{p_2} + \frac{p_2}{p_1} \right)^2$?