If the equation of a line parallel to $3x - 2y + 5 = 0$ and at a distance of $5$ units from it is $3x - 2y + C = 0$,then $C$ is equal to

Let $A(a, b)$,$B(3, 4)$,and $C(-6, -8)$ respectively denote the centroid,circumcentre,and orthocentre of a triangle. Then,the distance of the point $P(2a+3, 7b+5)$ from the line $2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ 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

The positions of the points $(3, 4)$ and $(2, -6)$ with respect to the line $3x - 4y = 8$ are:

If $p_{1}$ and $p_{2}$ are the lengths of perpendiculars from the origin to the lines $x \sin \theta + y \cos \theta = 5 \cos 2 \theta$ and $x \operatorname{cosec} \theta + y \sec \theta - 5 = 0$ respectively,then $p_{1}^{2} + 4 p_{2}^{2} = $

The line $L$ is given by $\frac{x}{5} + \frac{y}{b} = 1$ and 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