Lines are drawn parallel to the line $4x - 3y + 2 = 0$ at a distance of $\frac{3}{5}$ from the origin. Which one of the following points lies on any of these lines?

From the following options,the line nearest to the origin is ....

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 vertices of $\Delta OBC$ are $(0, 0)$,$(-3, 1)$,and $(-1, -3)$ respectively. $A$ line parallel to $BC$ intersects $OB$ and $OC$ at a distance of $1/2$ from $O$. Find the equation of this line.

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:

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: