The point of intersection of the diagonals of the rectangle whose sides are contained in the lines $x=8, x=10, y=11$ and $y=12$ is

$A$ triangle with vertices $(4, 0), (-1, -1), (3, 5)$ is

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

Three vertices of a triangle are $A(4, 3)$,$B(1, -1)$,and $C(7, k)$. The value$(s)$ of $k$ for which the centroid,orthocentre,incentre,and circumcentre of the $\Delta ABC$ lie on the same straight line is/are:

Let $A(h, k)$,$B(1, 1)$,and $C(2, 1)$ be the vertices of a right-angled triangle with $AC$ as its hypotenuse. If the area of the triangle is $1$ square unit,then the set of values which $k$ can take is given by

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 of the other diagonal is