Equations of diagonals of square formed by lines $x = 0,$ $y = 0,$$x = 1$ and $y = 1$are

If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x$ $\cos \theta+ y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$ between the coordinates axes, then $\alpha$ is equal to

A variable straight line passes through the points of intersection of the lines, $x + 2y = 1$ and $2x - y = 1$ and meets the co-ordinate axes in $A\,\, \&\,\, B$ . The locus of the middle point of $AB$ is :

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

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

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is