The length of the perpendicular from $(1, -2)$ to the line $12x + 5y + 63 = 0$ 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} = $

$A$ variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$,where $O$ is the origin,is:

The length of the perpendicular from the point $(b, a)$ to the line $\frac{x}{a} - \frac{y}{b} = 1$ is:

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

If the point $(1, a)$ lies between the straight lines $x + y = 1$ and $2(x + y) = 3$,then $a$ lies in the interval