Let $A(2, -3)$ and $B(-2, 1)$ be 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 the line

The locus of a point $P(x, y)$ which moves in such a way that the segment $OP$,where $O$ is the origin $(0, 0)$,has a slope of $\sqrt{3}$ is:

Let $a \neq 0, b \neq 0, c$ be three real numbers and $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}, \forall p, q \in \mathbb{R}$. If $L\left(\frac{2}{3}, \frac{1}{3}\right) + L\left(\frac{1}{3}, \frac{2}{3}\right) + L(2, 2) = 0$,then the line $ax + by + c = 0$ always passes through the fixed point:

$A$ rod of length $8$ units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$,respectively. If the locus of the point $P$,that divides the rod $AB$ internally in the ratio $2:1$ is $9(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76=0$,then $\alpha-\beta-\gamma$ is equal to :

Two points $A$ and $B$ with coordinates $(1, 1)$ and $(-2, 3)$ are given respectively. Then,the locus of a point $P$ such that the area of $\triangle PAB$ is $9 \text{ sq. units}$ is given by $......$

$A(2,3)$ and $B(3,-5)$ are two vertices of $\triangle ABC$. If the centroid of the $\triangle ABC$ moves on the line $2x+y-2=0$,then the locus of $C$ is