The locus of the point $P$ which is equidistant from $3x + 4y + 5 = 0$ and $9x + 12y + 7 = 0$ is:

The locus of the centroid of the triangle whose vertices are $(a \cos t, a \sin t)$,$(b \sin t, -b \cos t)$,and $(1, 0)$,where $t$ is a parameter,is:

If the sum of the distances of a point from two perpendicular lines in a plane is $1$,then its locus is:

Let $A=(2,3)$ and $B=(3,-5)$ be two vertices of $\triangle ABC$ such that $C$ is a point on the line $L \equiv 3x + 4y - 5 = 0$. Then the locus of the centroid of $\triangle ABC$ is a line parallel to

If a point $(x, y) \equiv (\tan \theta + \sin \theta, \tan \theta - \sin \theta)$,then the locus of $(x, y)$ is

$A$ particle is projected vertically upwards. If it has to stay above the ground for $12 \text{ seconds}$, then: