If the coordinates of the vertices of the triangle $ABC$ be $(-1, 6)$, $(-3, -9)$, and $(5, -8)$ respectively, then the equation of the median through $C$ is

The opposite vertices of a square are $(1, 2)$ and $(3, 8)$, then the equation of a diagonal of the square passing through the point $(1, 2)$, is

Let a triangle be bounded by the lines $L _{1}: 2 x +5 y =10$; $L _{2}:-4 x +3 y =12$ and the line $L _{3}$, which passes through the point $P (2,3)$, intersect $L _{2}$ at $A$ and $L _{1}$ at $B$. If the point $P$ divides the line-segment $A B$, internally in the ratio $1: 3$, then the area of the triangle is equal to

$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, - 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. 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 $3m/s$ and $2m/s$ respectively. Its locus is

The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a