The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:

If $\alpha = 1 + \sum_{r=1}^6 (-3)^{r-1} \binom{12}{2r-1}$,then the distance of the point $(12, \sqrt{3})$ from the line $\alpha x - \sqrt{3} y + 1 = 0$ is ..........

Find the equation of the line which is equidistant from the parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.

The equation of the base of an equilateral triangle is $12x+5y-65=0$. If one of its vertices is $(2,3)$,then the length of the side is

Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$. (in $units$)

Let $d$ be the distance between the parallel lines $3x - 2y + 5 = 0$ and $3x - 2y + 5 + 2\sqrt{13} = 0$. Let $L_1 \equiv 3x - 2y + k_1 = 0$ $(k_1 > 0)$ and $L_2 \equiv 3x - 2y + k_2 = 0$ $(k_2 > 0)$ be two lines that are at a distance of $\frac{4d}{\sqrt{13}}$ and $\frac{3d}{\sqrt{13}}$ from the line $3x - 2y + 5 = 0$,respectively. Then the combined equation of the lines $L_1 = 0$ and $L_2 = 0$ is: