Let $\alpha, \beta, \gamma, \delta \in \mathrm{Z}$ and let $\mathrm{A}(\alpha, \beta), \mathrm{B}(1,0), \mathrm{C}(\gamma, \delta)$ and $D(1,2)$ be the vertices of a parallelogram $\mathrm{ABCD}$. If $\mathrm{AB}=\sqrt{10}$ and the points $\mathrm{A}$ and $\mathrm{C}$ lie on the line $3 y=2 x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the  equation to one diagonal is $11x + 7y = 9$, then the equation to the other diagonal is :-

A point $P$ moves on the line $2x -3y + 4 = 0$. If $Q(1, 4)$ and $R(3, -2)$ are fixed points, then the locus of the centroid of $\Delta PQR$ is a line

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

A variable straight line passes through a fixed point $(a, b)$ intersecting the co-ordinates axes at $A\,\, \&\,\, B$. If $'O'$ is the origin then the locus of the centroid of the triangle $OAB$ is :

lf a line $L$ is perpendicular to the line $5x - y\,= 1$ , and the area of the triangle formed by the line $L$ and the coordinate axes is $5$, then the distance of line $L$ from the line $x + 5y\, = 0$ is