Find a composite transformation matrix that will reflect triangle ABC about the line (L) and also find its new co-ordinates.

Solution:

Equation of line is: $y=1 / 2 x+2$

Comparing with $y=m x+c, m=0.5 and c=2$

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

$\theta=26.56^{\circ}$

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

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

$\{\mathrm{tx}=0, \mathrm{ty}=-2\} \text { (-ve as direction is downward.) }$

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

$[\mathrm{R}] \quad=\left[\begin{array}{ccc}{\cos \theta} & {\sin \theta} & {0} \\ {-\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{array}\right]=\left[\begin{array}{ccc}{\cos (-26.56)} & {\sin (-26.56)} & {0} \\ {-\sin (-26.56)} & {\cos (-26.56)} & {0} \\ {0} & {0} & {1}\end{array}\right]$

$\therefore[\mathrm{R}] \quad=\left[\begin{array}{ccc}{0.89} & {-0.48} & {0} \\ {0.48} & {0.89} & {0} \\ {0} & {0} & {1}\end{array}\right]$

Step 3: Reflection of triangle about x-axis

Matrix for reflection about x-axis is given as,

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

Step 4: Inverse rotation of line to its original angle

$[\mathrm{Rx}]_{\mathrm{ccw}}^{-1}=\left[\begin{array}{ccc}{\cos \theta} & {\sin \theta} & {0} \\ {-\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{array}\right]=\left[\begin{array}{ccc}{\cos (26.56)} & {\sin (26.56)} & {0} \\ {-\sin (26.56)} & {\cos (26.56)} & {0} \\ {0} & {0} & {1}\end{array}\right]$

$=\left[\begin{array}{ccc}{0.89} & {0.48} & {0} \\ {-0.48} & {0.89} & {0} \\ {0} & {0} & {1}\end{array}\right]$

Step 5: Inverse Translation of Point P to its original position

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

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

Now,

The composite transformation matrix,

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

$=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {-2} & {1}\end{array}\right]\left[\begin{array}{ccc}{0.89} & {-0.48} & {0} \\ {0.48} & {0.89} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {-1} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{0.89} & {0.48} & {0} \\ {-0.48} & {0.89} & {0} \\ {0} & {0} & {1}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {2} & {1}\end{array}\right]$

$=\left[\begin{array}{ccc}{0.561} & {0.854} & {0} \\ {0.854} & {-0.56} & {0} \\ {-1.7} & {1.123} & {1}\end{array}\right]\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {2} & {1}\end{array}\right]$

$=\left[\begin{array}{ccc}{0.56} & {0.854} & {0} \\ {0.854} & {-0.56} & {0} \\ {-1.7} & {3.123} & {1}\end{array}\right]=\left[\begin{array}{ccc}{0.6} & {0.8} & {0} \\ {0.8} & {-0.6} & {0} \\ {-1.7} & {3.1} & {1}\end{array}\right]$

Now, new coordinate of triangle ABC are,

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

$=\left[\begin{array}{lll}{2} & {4} & {1} \\ {4} & {6} & {1} \\ {2} & {6} & {1}\end{array}\right]\left[\begin{array}{ccc}{0.6} & {0.8} & {0} \\ {0.8} & {-0.6} & {0} \\ {-1.7} & {3.1} & {1}\end{array}\right]$

$=\left[\begin{array}{lll}{2.7} & {2.3} & {1} \\ {5.5} & {2.7} & {1} \\ {4.3} & {1.1} & {1}\end{array}\right]$

$A^{\prime}=(2.7,2.3)$

$B^{\prime}=(5.5,2.7)$

$C^{\prime}=(4.3,1.1)$