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

The locus of the mid-points of the perpendiculars drawn from points on the line, $\mathrm{x}=2 \mathrm{y}$ to the line $\mathrm{x}=\mathrm{y}$ is 

Triangle formed by the lines $3x + y + 4 = 0$ , $3x + 4y -15 = 0$ and $24x -7y = 3$ is a/an

A straight line passing through $P(3, 1)$ meet the coordinates axes at $A$ and $B$. It is given that distance of this straight line from the origin $'O'$ is maximum. Area of triangle $OAB$ is equal to

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

The area of the triangle formed by the line $x\sin \alpha + y\cos \alpha = \sin 2\alpha $and the coordinates axes is