The locus of a variable point which forms a triangle of fixed area with two fixed points is

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 $......$

$A$ straight line passing through a fixed point $(3, 5)$ intersects the coordinate axes at two points $A$ and $B$. If the locus of $C(x, y)$,which forms a rectangle with the points $A, O$ (origin),and $B$,is $ax + 2hxy + by = 0$,then $a + b + h =$

If the line segment joining the points $(1,0)$ and $(0,1)$ subtends an angle of $45^{\circ}$ at a variable point $P$,then the equation of the locus of $P$ is

$A$ line cuts the $x$-axis at $A(7, 0)$ and the $y$-axis at $B(0, -5)$. $A$ variable line $PQ$ is drawn perpendicular to $AB$ cutting the $x$-axis at $P(a, 0)$ and the $y$-axis at $Q(0, b)$. If $AQ$ and $BP$ intersect at $R(h, k)$, the locus of $R$ 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: