Find the image of the point $(-1, 3)$ with respect to the line $x - y + 1 = 0$.

Sum of values of $k$ for which the image of the point $(\lambda^2 + 1, \lambda)$ with respect to the line $y = -3x + 6k$ is the point $(\lambda, \lambda - 1)$ is:

If the point $P(4,1)$ undergoes a reflection in the line $x-y=0$,then a translation through a distance of $2$ units along the positive $X$-axis and finally projected on the $X$-axis,then the coordinates of $P$ in the final position are

If $\alpha$ and $\beta$ are the angles made by the normals drawn from the origin to the lines $x+y+\sqrt{2}=0$ and $x-\sqrt{3}y-2=0$ with the positive direction of the $X$-axis respectively,measured in the anti-clockwise direction,then $\alpha+\beta=$

If the family of straight lines $ax + by + c = 0$,where $2a + 3b = 4c$,is concurrent at the point $P(l, m)$,then the foot of the perpendicular drawn from $P$ to the line $x + y + 1 = 0$ is

$A(3, -4)$ is a vertex of $\triangle ABC$ and $3x + 4y - 18 = 0$ is the perpendicular bisector of the side $AB$. If $C = (6, 3)$,then the centroid of the triangle is: