Two sides of a square are along the lines $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$. The area of the square is:

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is

The perpendicular distance from the point $(1, \pi)$ to the line joining $(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ (in polar coordinates) is

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ upon the straight line $y = x \tan \alpha + c$,where $c > 0$,is:

The vertices of a triangle $OBC$ are $(0,0)$,$(-3,-1)$,and $(-1,-3)$ respectively. The equation of the line parallel to $BC$ which is at a distance of $\frac{1}{2}$ unit from the origin and cuts $OB$ and $OC$ is: