If the straight lines $x + 3y = 4$,$3x + y = 4$,and $x + y = 0$ form a triangle,then the triangle 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

Let $ABC$ be an equilateral triangle,and let $KLMN$ be a rectangle with $K, L$ on $BC$,$M$ on $AC$,and $N$ on $AB$. Suppose $AN / NB = 2$ and the area of $\triangle BKN$ is $6$. The area of the $\triangle ABC$ 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

$A(0,2)$ and $C(6,4)$ are opposite vertices of square $ABCD$. The sum of the slopes of the sides passing through vertex $A$ 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: