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:

The number of points,having both coordinates as integers,that lie in the interior of the triangle with vertices $(0,0)$,$(0,41)$,and $(41,0)$ 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:

The incentre of the triangle formed by the lines $x+y=1$,$x=1$,and $y=1$ is

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

The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6 \text{ sq. units}$ with the coordinate axes,is