On which side of the line $7x + 5y - 9 = 0$ do the points $(3, 4)$ and $(-9, 6)$ lie?

The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:

The distance between the parallel lines $3x + 4y + 7 = 0$ and $3x + 4y - 5 = 0$ is

The points $(2,3)$ and $(-4,-4/3)$ lie on the opposite sides of the line $L \equiv 5x - 6y + k = 0$,where $k$ is an integer. If the points $(1,2)$ and $(4,5)$ lie on the same side of the line $L = 0$,then the perpendicular distance from the origin to the line $L = 0$ is:

Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point on the line $7x - 9y + 10 = 0$ to the lines $3x + 4y = 5$ and $12x + 5y = 7$, respectively. Then,

Identify the point on the line $2x + 3y + 7 = 0$,which is at a distance of $3$ units from $(1, -3)$.