If a line $L$ is perpendicular to the line $5x - y = 1$,and the area of the triangle formed by the line $L$ and the coordinate axes is $5$,then the distance of line $L$ from the line $x + 5y = 0$ is

Let the angles made with the positive x-axis by two straight lines drawn from the point $P(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_{1}$ and $\theta_{2}$. Then the value of $(\theta_{1}+\theta_{2})$ is:

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ 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:

The equation of the base of an equilateral triangle is $x + y = 2$ and the vertex is $(2, -1)$. The length of the side of the triangle is

If $12a + 5b = 9$,where $a, b \in \mathbb{R}$,then the minimum value of $a^2 + b^2$ is: