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

$A$ rectangle is formed by the lines $x=0, x=3, y=0$ and $y=4$. Let the line $L$ be perpendicular to $3x+y+6=0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, -5)$ from the line $L$ is equal to:

The distance of the point $(2, 5)$ from the line $3x + y + 4 = 0$,measured parallel to the line $3x - 4y + 8 = 0$,is

Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x - y = 0$.

If $(\lambda^2, \lambda+1), \lambda \in \mathbb{Z}$ belongs to the region between the lines $x+2y-5=0$ and $3x-y+1=0$ which includes the origin,then the possible number of such points is

$O(0,0), B(-3,-1), C(-1,-3)$ are vertices of a triangle $OBC$. $D$ is a point on $OC$ and $E$ is a point on $OB$. If the equation of $DE$ is $2x + 2y + \sqrt{2} = 0$,then the ratio in which the line $DE$ divides the altitude of the triangle $OBC$ from $O$ to $BC$ is: