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

If $p_{1}$ and $p_{2}$ are the lengths of perpendiculars from the origin to the lines $x \sin \theta + y \cos \theta = 5 \cos 2 \theta$ and $x \operatorname{cosec} \theta + y \sec \theta - 5 = 0$ respectively,then $p_{1}^{2} + 4 p_{2}^{2} = $

If $p$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$,then show that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$.

The length of the perpendicular drawn from the origin upon the straight line $\frac{x}{3} - \frac{y}{4} = 1$ is

The equation of the base of an equilateral triangle is $12x+5y-65=0$. If one of its vertices is $(2,3)$,then the length of the side is

Find the distance between the parallel lines $l(x + y) + p = 0$ and $l(x + y) - r = 0$.