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 $A (\cos \alpha, \sin \alpha)$,$B (\sin \alpha, -\cos \alpha)$,and $C (1, 2)$ are the vertices of $\Delta ABC$,find the locus of its centroid as $\alpha$ varies.

If $L : ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth,and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point:

Let $A$ be the set of all points $(\alpha, \beta)$ such that the area of the triangle formed by the points $(5, 6), (3, 2),$ and $(\alpha, \beta)$ is $12 \text{ square units}.$ Then the least possible length of a line segment joining the origin to a point in $A$ is:

The locus of a point which is at a distance of $2$ units from the line $2x - 3y + 4 = 0$ and at a distance of $\sqrt{13}$ units from the point $(5, 0)$ is:

$A$ variable line passes through a fixed point $(a, b)$ and meets the coordinate axes at $A$ and $B$. The locus of the point of intersection of lines drawn through $A$ and $B$ parallel to the coordinate axes is: