What type of triangle is formed by the lines $x + y = 0$,$3x + y = 4$,and $x + 3y = 4$?

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

The coordinates of the vertices of a quadrilateral are $(2, -1)$,$(0, 2)$,$(2, 3)$,and $(4, 0)$. The angle between its diagonals is:

The area of the quadrilateral formed by the lines $x+2y+3=0$, $2x+4y+9=0$, $x-2y+3=0$, and $3x-6y+11=0$ is

Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively. If $\text{Area}(PKAN) = 25$,$\text{Area}(PLBK) = 36$,and $\text{Area}(PMDN) = 41$,then $\text{Area}(PLCM)$ is

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