If the foot of the perpendicular drawn from the origin to a straight line is $(3, -4)$,then the equation of the line is:

The equation of the straight line passing through the point $(a \cos^3 \theta, a \sin^3 \theta)$ and perpendicular to the line $x \sec \theta + y \csc \theta = a$ is:

Find the coordinates of the foot of the perpendicular drawn from the point $(0, 5)$ to the line $3x - 4y - 5 = 0$.

The polar coordinates of $P$ are $\left(2, \frac{\pi}{6}\right)$. If $Q$ is the image of $P$ about the $X$-axis,then the polar coordinates of $Q$ are...

Let the mirror image of point $A(1, 4)$ in the line $y = x$ be point $B$; the mirror image of point $B$ in the line $y = -x$ be $C$; and the mirror image of $C$ in the $x$-axis be $D$. Then,the area of triangle $ABD$ is ............... $sq. \, units$.

The distance from the origin to the image of $(1,1)$ with respect to the line $x+y+5=0$ is (in $\sqrt{2}$)