$A$ 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:

If a point $(\alpha, \beta)$ on the line $3x + y = 0$ and the point $(3, 4)$ lie on opposite sides of the line $3x - 4y - 8 = 0$,then which of the following is correct?

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ 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 set of all possible values of $\theta$ in the interval $(0, \pi)$ for which the points $(1, 2)$ and $(\sin \theta, \cos \theta)$ lie on the same side of the line $x+y=1$ is

$A$ straight line makes an intercept on the $Y$-axis twice as long as that on the $X$-axis and is at a unit distance from the origin. Then the line is represented by the equations:

If the line $x-y+1=0$ cuts the lines $2x+2y+3=0$ and $3x+3y+2=0$ at the points $A$ and $B$ respectively,then $AB=$