Find the distance of the point $(3, -5)$ from the line $3x - 4y - 26 = 0$.

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 length of the perpendicular from the point $(b, a)$ to the line $\frac{x}{a} - \frac{y}{b} = 1$ is:

The number of lines that can be drawn through the point $(4, -5)$ at a distance of $10$ units from the point $(1, 3)$ is

Which pair of points lie on the same side of the line $3x - 8y - 7 = 0$?

The vertices of a triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to the origin is: