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What is the position of the point $(8, -9)$ with respect to the lines $2x + 3y - 4 = 0$ and $6x + 9y + 8 = 0$?

If the point $(1, a)$ lies between the lines $x + y = 1$ and $2(x + y) = 3$,then in which interval does $a$ lie?

Two sides of a square are along the lines $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$. The area of the square 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 distance between the lines $3x - 2y = 1$ and $6x - 4y + 9 = 0$ is

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