Let $\mathrm{C}$ be the centroid of the triangle with vertices $(3,-1),(1,3)$ and $(2,4) .$ Let $P$ be the point of intersection of the lines $x+3 y-1=0$ and $3 \mathrm{x}-\mathrm{y}+1=0 .$ Then the line passing through the points $\mathrm{C}$ and $\mathrm{P}$ also passes through the point

If vertices of a parallelogram are respectively $(0, 0)$, $(1, 0)$, $(2, 2)$ and $(1, 2)$, then angle between diagonals is 

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

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

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

The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a