If the perpendicular distance from the point $(1, 1)$ to the line $3x + 4y + c = 0$ is $7$,then the possible values of $c$ are:

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 equations of the lines passing through the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$ and whose distance from the point $(3, 2)$ is $\frac{7}{5}$ are:

$A$ variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$,where $O$ is the origin,is:

If $p$ and $q$ are the perpendicular distances from the origin to the straight lines $x \sec \theta - y \operatorname{cosec} \theta = a$ and $x \cos \theta + y \sin \theta = a \cos 2 \theta$,then

Find the length of the perpendicular drawn from the point $(2, 1)$ to the line $3x - 4y + 8 = 0$.