Let $A\ (2, -3)$ and $B\ (-2, 1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle moves on the line $2x + 3y = 1$,then the locus of the vertex $C$ is which line?

$A(a,0)$ and $B(-a,0)$ are two fixed points of triangle $ABC$. The vertex $C$ moves in such a way that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant. Then the locus of the point $C$ is

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

Given points $A(6,0)$,$B(0,4)$,and $O$ as the origin,find the locus of a point $P(x, y)$ such that the area of $\triangle POB$ is $2$ times the area of $\triangle POA$.

$A$ moving line intersects the lines $x+y=0$ and $x-y=0$ at the points $A$ and $B$ respectively, such that the area of the triangle with vertices $(0,0)$, $A$, and $B$ has a constant area $C$. The locus of the mid-point of $AB$ is given by the equation:

Consider the triangles with vertices $A(2,1)$,$B(0,0)$,and $C(t,4)$,where $t \in [0,4]$. If the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively,then $6\alpha + 21\beta$ is equal to $.........$.