The point whose abscissa is equal to its ordinate and which is equidistant from the points $(1, 0)$ and $(0, 3)$ is

$A$ straight line passes through a fixed point $(h, k)$. The locus of the foot of the perpendicular drawn from the origin to this line 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 equation of the locus of the foot of the perpendicular drawn from the origin to a line passing through a fixed point $(a, b)$ is:

$P(x, y)$ moves such that the area of the triangle formed by $P, Q(a, 2a)$ and $R(-a, -2a)$ is equal to the area of the triangle formed by $P, S(a, 2a)$ and $T(2a, 3a)$. The locus of $P$ is a straight line given by:

$A$ point starts moving from $(1, 2)$ and its projections on $x$ and $y$-axes are moving with velocities of $3 \ m/s$ and $2 \ m/s$ respectively. Its locus is