Find the composite transformation matrix and its new coordinates.

Solution:

Equation of line is $3 x-4 y+8=0$ i.e $4 y=3 x+8$ i.e $y=(3 / 4) x+2$

Comparing with $y=m x+c, m=3 / 4$ and $c=2$

also $\theta=\tan ^{-1} m$

$\theta=36.87^{\circ}$

Step 1: Transformation of Point $\mathrm{P}(0,2)$ to origin $\mathrm{O}(0,0)$

$[\operatorname{Tr}]=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {\operatorname{tx}} & {\operatorname{ty}} & {1}\end{array}\right]=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {-2} & {1}\end{array}\right]$

Step 2: Rotation of line by an angle of $\theta=-36.87^{\circ}$ (-ve as it is moving in clockwise direction).

$[\mathrm{R}]=\left[\begin{array}{ccc}{\cos \theta} & {\sin \theta} & {0} \\ {-\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{array}\right]=\left[\begin{array}{ccc}{\cos (-36.87)} & {\sin (-36.87)} & {0} \\ {-\sin (-36.87)} & {\cos (-36.87)} & {0} \\ {0} & {0} & {1}\end{array}\right]$

$\therefore[\mathrm{R}]=\left[\begin{array}{ccc}{0.8} & {-0.6} & {0} \\ {0.6} & {0.8} & {0} \\ {0} & {0} & {1}\end{array}\right]$

Step 3: Reflection of triangle about x-axis

Matrix for reflection about x-axis is given as

$[\mathrm{M}]_{@ \times-\mathrm{axis}}=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {-1} & {0} \\ {0} & {0} & {1}\end{array}\right]$

Step 4: Reverse rotation of line to its original angle

$[\mathrm{Rx}]_{\mathrm{ccw}}^{-1} \quad=\left[\begin{array}{ccc}{\cos \theta} & {\sin \theta} & {0} \\ {-\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{array}\right]=\left[\begin{array}{ccc}{\cos (36.87)} & {\sin (36.87)} & {0} \\ {-\sin (36.87)} & {\cos (36.87)} & {0} \\ {0} & {0} & {1}\end{array}\right]$

$=\left[\begin{array}{ccc}{0.8} & {0.6} & {0} \\ {-0.6} & {0.8} & {0} \\ {0} & {0} & {1}\end{array}\right]$

Step 5: Inverse Translation of Point $\mathrm{P}$ to its original position.

$[\mathrm{Tr}]^{-1}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {\operatorname{tx}} & {\text { ty }} & {1}\end{array}\right]=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {2} & {1}\end{array}\right]$

$\{\mathrm{tx}=0, \mathrm{ty}=+2\} \text { (+ve as direction is upwards) }$

Now, The composite transformation matrix

$[\mathrm{T}]=[\operatorname{Tr}][\mathrm{R}][\mathrm{M}][\mathrm{R}]^{-1}[\mathrm{Tr}]^{-1}$

$=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {-2} & {1}\end{array}\right]\left[\begin{array}{ccc}{0.8} & {-0.6} & {0} \\ {0.6} & {0.8} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {-1} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{0.8} & {0.6} & {0} \\ {-0.6} & {0.8} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {2} & {1}\end{array}\right]$

$=\left[\begin{array}{ccc}{0.28} & {0.96} & {0} \\ {0.96} & {-0.28} & {0} \\ {-1.92} & {0.56} & {1}\end{array}\right]\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {2} & {1}\end{array}\right]$

$[\mathrm{T}]=\left[\begin{array}{ccc}{0.28} & {0.96} & {0} \\ {0.96} & {-0.28} & {0} \\ {-1.92} & {2.56} & {1}\end{array}\right]$

Now, new coordinate of triangle ABC are

$\left[\mathrm{X}^{\prime}\right]=[\mathrm{X}][\mathrm{T}]$

$=\left[\begin{array}{lll}{4} & {1} & {1} \\ {5} & {2} & {1} \\ {4} & {3} & {1}\end{array}\right]\left[\begin{array}{ccc}{0.28} & {0.96} & {0} \\ {0.96} & {-0.28} & {0} \\ {-1.92} & {2.56} & {1}\end{array}\right]$

$=\left[\begin{array}{ccc}{0.16} & {6.12} & {1} \\ {1.4} & {6.8} & {1} \\ {2.08} & {5.56} & {1}\end{array}\right]$

$A^{\prime}=(0.16,6.12)$

$B^{\prime}=(1.4,6.8)$

$C^{\prime}=(2.08,5.56)$