The reflection of the point $(4, -13)$ in the line $5x + y + 6 = 0$ is

Assuming that straight lines work as the plane mirror for a point,find the image of the point $(1, 2)$ in the line $x - 3y + 4 = 0$.

Let the mirror image of point $A(1, 4)$ in the line $y = x$ be point $B$; the mirror image of point $B$ in the line $y = -x$ be $C$; and the mirror image of $C$ in the $x$-axis be $D$. Then,the area of triangle $ABD$ is ............... $sq. \, units$.

If $2x + 3y = 5$ is the perpendicular bisector of the line segment joining the points $A\left(1, \frac{1}{3}\right)$ and $B$,then $B$ is equal to

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 . . . . . . .

The coordinates of the image point $Q$ of the point $P(-5, 13)$ with respect to the line $2x - 3y - 3 = 0$ are: