If the ratio of the distances of a variable point $P$ from the point $(1,1)$ and the line $x-y+2=0$ is $1: \sqrt{2}$, then the equation of the locus of $P$ is

If the origin is shifted to the point $\left(\frac{3}{2},-2\right)$ by the translation of axes, then the transformed equation of $2 x^2+4 x y+y^2+2 x-2 y+1=0$ is

$L \equiv x \cos \alpha+y \sin \alpha-p=0$ represents a line perpendicular to the line $x+y+1=0$. If $p$ is positive, $\alpha$ lies in the fourth quadrant and perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line, $L=0$ is 5 units, then $p=$

$(-2,-1),(2,5)$ are two vertices of a triangle and $\left(2, \frac{5}{3}\right)$ is its orthocenter. If $(m, n)$ is the third vertex of that triangle, then $m+n$ is equal to.