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

Let $A(1, 1)$ be a point. $B$ is the image of $A$ with respect to the line $x + 2y + 2 = 0$. If $C$ is the foot of the perpendicular from $B$ on the line $3x + 4y - 10 = 0$,then $AC$ is equal to

$A$ person standing at the junction (crossing) of $2$ straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$,wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. The equation of the path he should follow is:

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 ............... $sq. \, units$.

$A$ light beam emanating from the point $A(3, 10)$ reflects from the line $2x + y - 6 = 0$ and then passes through the point $B(5, 6)$. The equations of the incident and reflected beams are respectively:

Let $\alpha$ and $\beta$ be integers satisfying $0 < \beta < \alpha$. Let $P(\alpha, \beta)$ be a point. Let $Q$ be the reflection of $P$ in the line $y = x$,$R$ be the reflection of $Q$ in the $y$-axis,$S$ be the reflection of $R$ in the $x$-axis,and $T$ be the reflection of $S$ in the $y$-axis. If the area of the convex pentagon $PQRST$ is $187 \ sq. \ units$,then the value of $\alpha + \beta^2$ is: