The locus of the orthocentre of the triangle formed by the lines

$ (1+p) x-p y+p(1+p)=0, $

$ (1+q) x-q y+q(1+q)=0,$

and $y=0$, where $p \neq q$, 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

Area of the parallelogram whose sides are $x\cos \alpha + y\sin \alpha = p$ $x\cos \alpha + y\sin \alpha = q,\,\,$ $x\cos \beta + y\sin \beta = r$ and $x\cos \beta + y\sin \beta = s$ 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

A vertex of square is $(3, 4)$ and diagonal $x + 2y = 1,$ then the second diagonal which passes through given vertex will be

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations to the sides $AB$ and $AC$ are respectively $px+qy= 1$ and $qx + py = 1.$ Then the equation to the median through $A$ is