Jika titik M (3,4) dirotasikan berturut-turut terhadap [P (2,1), R1 (10°)] dan [P (2,1), R2D (35°)], bayangannya adalah..

jawab
MT =
[tex] \dbinom{cos \: (10 + 35) \: \: \: \: \: \: \: \: - sin \: (10 + 35)}{sin \: (10 + 35) \: \: \: \: \: \: \: \: \: cos \: (10 + 35)} \\ \\ = \dbinom{cos \: 45 \: \: \: \: \: - sin \: 45}{sin \: 45 \: \: \: \: \: \: \: cos \: 45} = \dbinom{ \dfrac{1}{2} \sqrt{2} \: \: \: \: \: \: \: \times - \dfrac { 1}{2} \sqrt{2} }{ \dfrac{1}{2} \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{1}{2} \sqrt{2} } \\ [/tex]

(x', y') = MT × (x, y)

[tex] \dbinom{ {x}^{l} }{ {y}^{l} } = \dbinom{ \dfrac{1}{2} \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: - \dfrac{1}{2} \sqrt{2} }{ \dfrac{1}{2} \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{1}{2} \sqrt{2} } \dbinom{3 - 2}{4 - 1} + \dbinom{2}{1} \\ \dbinom{ {x}^{l} }{ {y}^{l} } = \dbinom{ \dfrac{1}{2} \sqrt{2} \: \: \: \: - \dfrac{1}{2}\sqrt{2} }{ \dfrac{1}{2} \sqrt{2} \: \: \: \: \dfrac{1}{2}\sqrt{ 2} } \dbinom{1}{3}+ \dbinom{2}{1} \\ \dbinom{ {x}^{l} }{ {y}^{l} } = \dbinom{ \dfrac {1}{2} \sqrt{2} - \dfrac {3}{2} \sqrt{2} } { \dfrac {1}{2} \sqrt{2} + \dfrac {3}{2} \sqrt{2}} + \dbinom{2}{1} \\
\dbinom{ {x}^{l} }{ {y}^{l} } = \dbinom{ -1 \sqrt{2}} {2 \sqrt{2}}+ \dbinom{2}{1} [/tex]

maka titik bayanganya (-√2 + 2, 2√2 + 1) atau (2 - √2, 1 + 2√2)