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

Let the angles made with the positive x-axis by two straight lines drawn from the point $P(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_{1}$ and $\theta_{2}$. Then the value of $(\theta_{1}+\theta_{2})$ is:

Given the lines $y + 2x = 3$ and $y + 2x = 5$ cut the axes at $A, B$ and $C, D$ respectively.
Statement-$1$ : $ABDC$ forms a quadrilateral and the point $(2, 3)$ lies inside the quadrilateral.
Statement-$2$ : The point $(2, 3)$ lies between the two parallel lines.

Find the distance between the lines $3x + 4y + 7 = 0$ and $3x + 4y + 22 = 0$.

$A$ 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

Let the expression $E = 8^a + 8^b - 3 \cdot 2^{a+b}$ take its minimum value $p$ at $a = \alpha$ and $b = \beta$. Then,the perpendicular distance of the point $P(\alpha, \beta)$ from the line $x + y + 2p = 0$ is