If the equation of base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$, then the length of the side of the triangle is

Two sides of a parallelogram are along the lines, $x + y = 3$ and $x -y + 3 = 0$. If its diagonals intersect at $(2, 4)$, then one of its vertex is

Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side a such that $a^{2}+11 a+3\left(m_{2}^{2}+m_{2}^{2}\right)=220$. If one vertex of the square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$, where $\alpha \in\left(0, \frac{\pi}{2}\right)$ and the equation of one diagonal is $(\cos \alpha-\sin \alpha) x +(\sin \alpha+\cos \alpha) y =10$, then  $72 \left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$ is equal to.

The area of the parallelogram formed by the lines $y = mx,\,y = mx + 1,\,y = nx$ and $y = nx + 1$ equals

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

If the vertices $P$ and $Q$ of a triangle $PQR$ are given by $(2, 5)$ and $(4, -11)$ respectively, and the point $R$ moves along the line $N: 9x + 7y + 4 = 0$, then the locus of the centroid of the triangle $PQR$ is a straight line parallel to