The coordinates of the two points lying on $x+y=4$ and at a unit distance from the straight line $4x+3y=10$ are

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 distance between two parallel lines $3x + 4y - 8 = 0$ and $3x + 4y - 3 = 0$ is given by

The vertices of a triangle $OBC$ are $(0,0)$,$(-3,-1)$,and $(-1,-3)$ respectively. The equation of the line parallel to $BC$ which is at a distance of $\frac{1}{2}$ unit from the origin and cuts $OB$ and $OC$ 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

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