If the distance from a variable point $P$ to the point $(4, 3)$ is equal to the perpendicular distance from $P$ to the line $x + 2y - 1 = 0$,then the equation of the locus of the point $P$ is

$A$ $10 \text{ inches}$ long pencil $AB$ with midpoint $C$ and a small eraser $P$ are placed on the horizontal top of a table such that $PC = \sqrt{5} \text{ inches}$ and $\angle PCB = \tan^{-1}(2)$. The acute angle through which the pencil must be rotated about $C$ so that the perpendicular distance between the eraser and the pencil becomes exactly $1 \text{ inch}$ is:

The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is

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:

If $A=(5,3)$,$B=(3,-2)$ and a point $P$ is such that the area of the triangle $PAB$ is $9$,then the locus of $P$ represents

$A$ straight line meets the $X$ and $Y$ axes at the points $A$ and $B$ respectively. If $AB = 6$ units,then the locus of the point $P$ which divides the line segment $AB$ such that $AP : PB = 2 : 1$ is