The acute angle between the diagonals of a parallelogram whose vertices are $A(2, -1), B(0, 2), C(2, 3)$ and $D(4, 0)$ is

If the points $(1, 1)$,$(-1, -1)$,and $(-\sqrt{3}, \sqrt{3})$ are the vertices of a triangle,then this triangle is:

Consider a $\triangle ABC$ in the $XY$-plane with vertices $A=(0,0)$,$B=(1,1)$,and $C=(9,1)$. If the line $x=a$ divides the triangle into two parts of equal area,then $a$ equals

The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

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

Let the area of the triangle with vertices $A(1, \alpha)$,$B(\alpha, 0)$,and $C(0, \alpha)$ be $4 \text{ sq. units}$. If the points $(\alpha, -\alpha)$,$(-\alpha, \alpha)$,and $(\alpha^2, \beta)$ are collinear,then $\beta$ is equal to: