Let $A$ be the set of all points $(\alpha, \beta)$ such that the area of the triangle formed by the points $(5, 6), (3, 2),$ and $(\alpha, \beta)$ is $12 \text{ square units}.$ Then the least possible length of a line segment joining the origin to a point in $A$ 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:

$A$ line passing through the point $(1, 2)$ meets the axes at $P$ and $Q$ such that it forms a triangle $OPQ$,where $O$ is the origin. If the area of triangle $OPQ$ is minimum,then the slope of the line $PQ$ is:

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA$ is equal to:

$A$ light ray emerging from the point source placed at $P(1, 3)$ is reflected at a point $Q$ on the $x$-axis. If the reflected ray passes through the point $R(6, 7)$,then the abscissa of $Q$ is

For a variable line $\frac{x}{a} + \frac{y}{b} = 1$ where $a + b = 10$,find the equation of the locus of the midpoint of the portion of the line intercepted between the coordinate axes.