What are the coordinates of the image of the point $(4, -5)$ with respect to the line $y = x$?

From the point $(-1, -1)$, two rays are sent making angles of $45^\circ$ with the line $x+y=0$. These rays get reflected from the mirror $x+2y=1$. If the equations of the reflected rays are $ax+by=9$ and $cx+dy=7$, where $a, b, c, d \in Z$, then the value of $ad+bc$ is . . . . . . .

Let the triangle $PQR$ be the image of the triangle with vertices $(1,3), (3,1)$ and $(2,4)$ in the line $x+2y=2$. If the centroid of $\triangle PQR$ is the point $(\alpha, \beta)$,then $15(\alpha-\beta)$ is equal to:

The equation of the perpendicular bisector of the line segment joining the points $(7, 4)$ and $(-1, -2)$ is:

If the normal drawn from the origin to the straight line $2x + 7y + 6 = 0$ makes an angle $\theta$ with the positive $X$-axis,then $\theta =$

If $(h, k)$ is the image of the point $(3, -4)$ with respect to the line $2x - 3y - 5 = 0$ and $(\ell, m)$ is the foot of the perpendicular from $(h, k)$ onto the line $3x + 2y + 12 = 0$,then $\ell h + mk + 1 =$ ?