The distance between two parallel lines $3x + 4y - 8 = 0$ and $3x + 4y - 3 = 0$ is given by

If $p$ and $p'$ are the distances of the origin from the lines $x \sec \alpha + y \csc \alpha = k$ and $x \cos \alpha - y \sin \alpha = k \cos 2\alpha$,then $4p^2 + p'^2$ equals:

If the line $2x - y + 3 = 0$ is at a distance of $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$ from the lines $4x - 2y + \alpha = 0$ and $6x - 3y + \beta = 0$ respectively,then the sum of all possible values of $\alpha$ and $\beta$ is:

For the three points $A(2,0)$,$B(0,2)$,and $P(1,1)$,suppose $d$ is the algebraic sum of the distances of $A$ and $B$ from a line that passes through $P$. Then,which of the following is correct?

$A$ 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:

The points $(2,3)$ and $(-4,-4/3)$ lie on the opposite sides of the line $L \equiv 5x - 6y + k = 0$,where $k$ is an integer. If the points $(1,2)$ and $(4,5)$ lie on the same side of the line $L = 0$,then the perpendicular distance from the origin to the line $L = 0$ is: