$P$ is a point on $x+y+5=0$,whose perpendicular distance from $2x+3y+3=0$ is $\sqrt{13}$. Then the coordinates of $P$ are:

Let $A(a, b)$,$B(3, 4)$,and $C(-6, -8)$ respectively denote the centroid,circumcentre,and orthocentre of a triangle. Then,the distance of the point $P(2a+3, 7b+5)$ from the line $2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ is

Let the origin be the centroid of an equilateral triangle $ABC$ and one of its sides be along the straight line $x+y=3$. If $R$ and $r$ are its circumradius and inradius respectively,then $R+r=$

The equation of the line passing through $(1, 2)$ and having a distance equal to $7$ units from the point $(8, 9)$ is:

Statement $(A)$: The points $(2, 1)$ and $(-3, 5)$ lie on opposite sides of the line $3x - 2y + 1 = 0$.
Reason $(R)$: The algebraic perpendicular distances from the given points to the line have opposite signs.

For an integer $K$,if the point $P(K^2, K+1)$ and the origin $O(0,0)$ lie in the same region between the lines $x+2y-5=0$ and $3x-y+1=0$,then the possible number of such points $P$ is