Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$. (in $/5$)

Find the set of all values of $a$ such that both the points $(1, 2)$ and $(3, 4)$ lie on the same side of the line $3x - 5y + a = 0$.

Two sides of a square are along the lines $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$. The area of the square is:

If the perpendicular distances from the points $(2, 3)$,$(4, a)$ and $(\alpha, \beta)$ to the line $3x + 4y - 3 = 0$ are equal and $4\alpha - 3\beta + 1 = 0$,then the sum of all possible values of $a$,$\alpha$,and $\beta$ is:

Statement-$1$: There is one line through $A(4, -5)$ such that its distance from $B(-2, 3)$ is $12$.
Statement-$2$: $AB = 10$.

The coordinates of the points on the line $2x - y = 5$ which are at a distance of $1$ unit from the line $3x + 4y = 5$ are: