If the points $(1, 2)$ and $(3, 4)$ lie on the same side of the straight line $3x - 5y + a = 0$,then $a$ lies in the set

The system of equations $x+2y=3$ and $3x+6y=a-2$ has no solution if:

Reduce the equation $x-y=4$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.

If $x + 2y = 3$ is a line and $A(-1, 3)$,$B(2, -3)$,and $C(4, 9)$ are three points,then:

$A$ line $L_1$ passing through the point of intersection of the lines $x-2y+3=0$ and $2x-y=0$ is parallel to the line $L_2$. If $L_2$ passes through the origin and also through the point of intersection of the lines $3x-y+2=0$ and $x-3y-2=0$,then the distance between the lines $L_1$ and $L_2$ is

If $12a + 5b = 9$,where $a, b \in \mathbb{R}$,then the minimum value of $a^2 + b^2$ is: