Find the distance between the lines $5x + 12y + 13 = 0$ and $5x + 12y - 9 = 0$.

If $(\sin \theta, \cos \theta)$ and $(3, 2)$ lie on the same side of the line $x + y = 1$,then $\theta$ lies in the interval:

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:

The length of the intercept of the line $x+1=0$ between the lines $3x+2y=5$ and $3x+2y=3$ is

$A$ rectangle is formed by the lines $x=0, x=3, y=0$ and $y=4$. Let the line $L$ be perpendicular to $3x+y+6=0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, -5)$ from the line $L$ is equal to:

The number of lines which pass through the point $(2, -3)$ and are at a distance of $8$ from the point $(-1, 2)$ is: