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 the line $x+y=1$ meet the axes of $x$ and $y$ at $A$ and $B$,respectively. $A$ right-angled triangle $AMN$ is inscribed in the triangle $OAB$,where $O$ is the origin and the points $M$ and $N$ lie on the lines $OB$ and $AB$,respectively. If the area of the triangle $AMN$ is $\frac{4}{9}$ of the area of the triangle $OAB$ and $AN : NB = \lambda : 1$,then the sum of all possible values of $\lambda$ is:

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

The equations of the sides of a triangle are $x - 3y = 0$,$4x + 3y = 5$,and $3x + y = 0$. The line $3x - 4y = 0$ passes through:

Let $A(-2,-1)$,$B(1,0)$,$C(\alpha, \beta)$,and $D(\gamma, \delta)$ be the vertices of a parallelogram $ABCD$. If the point $C$ lies on $2x-y=5$ and the point $D$ lies on $3x-2y=6$,then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to . . . . . . .

The area of the triangle enclosed by the straight lines $x = 0$,$y = 0$ and $x + 2y + 3 = 0$ in sq. units is: