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

Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a^2$ is equal to

$x+8y-22=0$, $5x+2y-34=0$, and $2x-3y+13=0$ are the three sides of a triangle. The area of the triangle is

The area of the parallelogram formed by the lines $L_1 \equiv \lambda x+4 y+2=0$,$L_2 \equiv 3 x+4 y-3=0$,$L_3 \equiv 2 x+\mu y+6=0$,and $L_4 \equiv 2 x+y+3=0$,where $L_1$ is parallel to $L_2$ and $L_3$ is parallel to $L_4$,is

$\Delta ABC$ has integral side lengths. The internal angle bisector of angle $B$ meets side $AC$ at $D$. If $AD = 3$ and $DC = 8$,then the perimeter of $\Delta ABC$ is:

Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. An equilateral triangle $ABP$ is constructed on the side of $AB$ remote from the origin. The orthocentre of triangle $ABP$ is: