Find the equation of a line passing through the point $(4, 3)$,which cuts a triangle of minimum area from the first quadrant.

The locus of a point $P$ which moves such that the sum of its distances from two perpendicular lines is equal to $1$ is a

The locus of the centroid of a triangle whose vertices are $(1, 0)$,$(a \cos t, a \sin t)$,and $(b \sin t, -b \cos t)$ is $9x^2 + 9y^2 - 6x = k$. Then,the value of $k$ is equal to

The point $P(2, 1)$ is translated to a point $Q$ parallel to the line $L \equiv x-y-4=0$ by $2 \sqrt{3}$ units. If the point $Q$ lies in the third quadrant,then the equation of the line passing through $Q$ and perpendicular to $L$ is

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA$ is equal to:

$A$ point starts moving from $(1, 2)$ and its projections on $x$ and $y$-axes are moving with velocities of $3 \ m/s$ and $2 \ m/s$ respectively. Its locus is