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

Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are

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

Locus of the points which are at equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is

If one vertex of an equilateral triangle of side $'a'$ lies at the origin and the other lies on the line $x - \sqrt{3} y = 0$ then the co-ordinates of the third vertex are :

Let $\alpha, \beta, \gamma, \delta \in \mathrm{Z}$ and let $\mathrm{A}(\alpha, \beta), \mathrm{B}(1,0), \mathrm{C}(\gamma, \delta)$ and $D(1,2)$ be the vertices of a parallelogram $\mathrm{ABCD}$. If $\mathrm{AB}=\sqrt{10}$ and the points $\mathrm{A}$ and $\mathrm{C}$ lie on the line $3 y=2 x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to