The distance between the lines given by $3x + 4y = 9$ and $6x + 8y = 15$ is (in $units$)

If $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line $x+y+3=0$ such that each of them is at a distance of $\sqrt{5}$ units from the line $x+2y+2=0$,then the value of $|x_1-x_2|$ 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:

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:

Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point on the line $7x - 9y + 10 = 0$ to the lines $3x + 4y = 5$ and $12x + 5y = 7$, respectively. Then,

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ to the line $y = x \tan \alpha + c, c > 0$ is .....