The diagonal passing through origin of a quadrilateral formed by $x = 0,\;y = 0,\;x + y = 1$ and $6x + y = 3,$ is

The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

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 :-

The four points whose co-ordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form :

The equation of the base of an equilateral triangle is $x + y = 2$ and the vertex is $(2, -1)$. The length of the side of the triangle is

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