$A(2,3)$ 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

$A$ straight line passing through a point $(3, 2)$ cuts the $X$ and $Y$-axes at points $A$ and $B$ respectively. If a point $P(h, k)$ divides $AB$ in the ratio $2: 3$,then the equation of the locus of point $P$ is

$A$ point equidistant from the points $(2, 0)$ and $(0, 2)$ is

$A$ variable straight line passes through the point of intersection of the lines $x + 2y = 1$ and $2x - y = 1$ and meets the coordinate axes at $A$ and $B$. The locus of the midpoint of $AB$ is:

$A$ variable line passes through a fixed point $P$. If the algebraic sum of the perpendiculars drawn from $(2, 0)$,$(0, 2)$,and $(1, 1)$ to the line is zero,then the coordinates of the point $P$ are:

Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.