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

Two sides of a parallelogram are along the lines $4 x+5 y=0$ and $7 x+2 y=0$. If the equation of one of the diagonals of the parallelogram is $11 \mathrm{x}+7 \mathrm{y}=9$, then other diagonal passes through the point:

Let the points $\left(\frac{11}{2}, \alpha\right)$ lie on or inside the triangle with sides $x + y =11, x +2 y =16$ and $2 x +3 y =29$. Then the product of the smallest and the largest values of $\alpha$ is equal to :

Let ${ }^n C_{r-1}=28,{ }^n C_r=56$ and ${ }^n C_{r+1}=70$. Let $A (4 \cos t, 4 \sin t ), B (2 \sin t ,-2 \cos t )$ and $C \left(3 r - n , r ^2- n -1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3 x-1)^2+(3 y)^2=\alpha$, is the locus of the centroid of triangle $ABC$ , then $\alpha$ equals :

A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$, respectively. If the locus of the point $P$, that divides the $\operatorname{rod} AB$ internally in the ratio $2: 1$ is $9\left(x^2+\alpha y^2+\beta x y+\gamma x+28 y\right)-76=0$, then $\alpha-\beta-\gamma$ is equal to :