Determine the new vertex position if.

i) The triangle is rotated by $60^0$, anticlockwise about the vertex A.

ii) The triangle is scale by 2 times in X direction and 3 times in Y direction about vertex A.

Solution:

A) The triangle is rotated by $60^0$, anticlockwise about the vertex A.

Tr = Translate point $(-5,-5)$ to origin $\mathrm{O}(0,0)$

$\mathrm{R}=$ Rotate by 60 deg. Anticlockwise.

$Tr ^{-1}=$ Inverse Translation to original position a $(5,5)$

$$[\mathrm{T}]=(\mathrm{Tr})(\mathrm{R})(T)^{-1}$$

$$[\mathrm{T}]=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {t x} & {t y} & {1}\end{array}\right] \times\left[\begin{array}{ccc}{\cos 60} & {\sin 60} & {0} \\ {-\sin 60} & {\cos 60} & {0} \\ {0} & {0} & {1}\end{array}\right] \times\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {t x} & {t y} & {1}\end{array}\right]$$

$$[\mathrm{T}]=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {-5} & {-5} & {1}\end{array}\right] \times\left[\begin{array}{ccc}{\cos 60} & {\sin 60} & {0} \\ {-\sin 60} & {\cos 60} & {0} \\ {0} & {0} & {1}\end{array}\right] \times\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {5} & {5} & {1}\end{array}\right]$$

$$[\mathrm{T}]=\left[\begin{array}{ccc}{0.5} & {0.86} & {0} \\ {-0.86} & {0.5} & {0} \\ {-6.83} & {1.83} & {1}\end{array}\right]$$

$$[\mathrm{X}]=[\mathrm{X}] \times [\mathrm{T}]$$

$$\left[\begin{array}{ccc}{5} & {5} & {0} \\ {8} & {5} & {0} \\ {5} & {10} & {0}\end{array}\right] \times\left[\begin{array}{ccc}{0.5} & {0.86} & {0} \\ {-0.86} & {0.5} & {0} \\ {6.83} & {-1.83} & {1}\end{array}\right]$$

$$[\mathrm{X}]=\left[\begin{array}{lll}{5.0} & {5.0} & {1} \\ {6.5} & {7.6} & {1} \\ {0.67} & {7.5} & {1}\end{array}\right]$$

$$A=5,5$$ $$B=6.5,7.6$$ $$C=0.67,7.5$$

B) The triangle is scale by 2 times in X direction and 3 times in Y direction about vertex A.

$\mathrm{Tr}=$ Translate point $(-5,-5)$ to origin $\mathrm{O}(0,0)$

$\mathrm{Sc}=$ scaling in $\mathrm{x}$ and $\mathrm{y}$ direction.

$T r^{-1}=$ Inverse Translation to original position a $(5,5)$

$$[\mathrm{T}]=(\mathrm{Tr})(\mathrm{Sc})(T)^{-1}$$

$$[\mathrm{T}]=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {t x} & {t y} & {1}\end{array}\right] X\left[\begin{array}{ccc}{S x} & {0} & {0} \\ {0} & {S y} & {0} \\ {0} & {0} & {1}\end{array}\right] X\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {t x} & {t y} & {1}\end{array}\right]$$

$$[\mathrm{T}]=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {-5} & {-5} & {1}\end{array}\right] \times\left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {3} & {0} \\ {0} & {0} & {1}\end{array}\right] \times \left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {5} & {5} & {1}\end{array}\right]$$

$$[\mathrm{T}]=\left[\begin{array}{lll}{2} & {0} & {0} \\ {0} & {3} & {0} \\ {5} & {10} & {1}\end{array}\right]$$

$$[\mathrm{X}]=[\mathrm{X}] \mathrm{x}[\mathrm{T}]$$

$$\left[\begin{array}{ccc}{5} & {5} & {0} \\ {8} & {5} & {0} \\ {5} & {10} & {0}\end{array}\right] \times\left[\begin{array}{ccc}{2} & {0} & {0} \\ {0} & {3} & {0} \\ {5} & {10} & {1}\end{array}\right]$$

$$[\mathrm{X}]=\left[\begin{array}{ccc}{10} & {15} & {0} \\ {16} & {15} & {0} \\ {10} & {30} & {0}\end{array}\right]$$

$$\begin{aligned} A &=10,15 \\ B &=16,15 \\ C &= 10,30 \end{aligned}$$