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

The system of equations $x + y = 2$ and $2x + 2y = 3$ will have:

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

Reduce the equation $x-\sqrt{3} y+8=0$ into normal form. Find the perpendicular distance from the origin and the angle between the perpendicular and the positive $x$-axis.

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is

Let two points be $A(1, -1)$ and $B(0, 2)$. If a point $P(x', y')$ is such that the area of $\Delta PAB = 5 \; \text{sq units}$ and it lies on the line $3x + y - 4\lambda = 0$,then a value of $\lambda$ is