The vertex of the right angle of a right-angled triangle lies on the straight line $2x + y - 10 = 0$. If the two other vertices are at points $(2, -3)$ and $(4, 1)$,then the area of the triangle in sq. units is:

Let the points of intersection of the lines $x-y+1=0$,$x-2y+3=0$,and $2x-5y+11=0$ be the midpoints of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ 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:

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the equation to one diagonal is $11x + 7y = 9$,then the equation to the other diagonal is:

Area of the rhombus bounded by the four lines,$ax \pm by \pm c = 0$ is :

Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side $a$ such that $a^{2}+11 a+3(m_{1}^{2}+m_{2}^{2})=220$. If one vertex of the square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$,where $\alpha \in(0, \frac{\pi}{2})$ and the equation of one diagonal is $(\cos \alpha-\sin \alpha) x +(\sin \alpha+\cos \alpha) y =10$,then $72(\sin ^{4} \alpha+\cos ^{4} \alpha)+a^{2}-3 a+13$ is equal to.