The angle made by a line $L$ with the positive $X$-axis measured in the positive direction is $\frac{\pi}{6}$ and the intercept made by $L$ on the $Y$-axis is negative. If $L$ is at a distance of $5$ units from the origin,then the perpendicular distance from the point $(1, -\sqrt{3})$ to the line $L$ is

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 . . . . . . .

What are the points on the $y$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units?

The distance of the point $(-2, 3)$ from the line $x - y - 5 = 0$ is

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:

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 ..........