The distance between the lines given by $3x + 4y = 9$ and $6x + 8y = 15$ is (in $units$)

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the straight lines $x \sec \alpha + y \operatorname{cosec} \alpha = 10$ and $x \cos \alpha - y \sin \alpha = 10 \cos 2 \alpha$,then $4 p^2 + q^2 =$

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:

The vertices of a triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to the origin is:

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:

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ upon the straight line $y = x \tan \alpha + c$,where $c > 0$,is: