A point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is

Co-ordinates of the orthocentre of the triangle whose vertices are $A(0, 0) , B(3, 4)$ and $C(4, 0)$ 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

A point starts moving from $(1, 2)$ and its projections on $x$ and $y$ - axes are moving with velocities of $3m/s$ and $2m/s$ respectively. Its locus is

If in triangle $ABC$ ,$ A \equiv (1, 10) $, circumcentre $\equiv$ $\left( { - \,\,{\textstyle{1 \over 3}}\,\,,\,\,{\textstyle{2 \over 3}}} \right)$ and orthocentre $\equiv$ $\left( {{\textstyle{{11} \over 3}}\,\,,\,\,{\textstyle{4 \over 3}}} \right)$ then the co-ordinates of mid-point of side opposite to $A$ is :

Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is