$A$ pair of straight lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ form a square. The coordinates of the centre of the circle inscribed in the square are

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 of the other diagonal is

If $ad-bc \neq 0$,then the area (in sq. units) of the parallelogram formed by the lines $ax+by+2=0$,$ax+by+5=0$,$cx+dy+3=0$ and $cx+dy+7=0$ 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.

The area of the triangle formed by the lines $x = 0$,$y = 0$,and $x/a + y/b = 1$ is:

The four points whose coordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form: