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Question-64309




Question Number 64309 by aseer imad last updated on 16/Jul/19
Commented by Tony Lin last updated on 17/Jul/19
let A′=A−(1,3)=(0,0),C′=C−(1,3)=(4,5)  D′⇒((√2)/2)(4+5i)(cos(π/4)+isin(π/4))=−(1/2)+(9/2)i  ⇒D=(1,3)+(−(1/2),(9/2))=((1/2),((15)/2))  B′⇒((√2)/2)(4+5i)(cos(π/4)−isin(π/4))=(9/2)+(1/2)i  ⇒B=(1,3)+((9/2),(1/2))=(((11)/2),(7/2))
$${let}\:{A}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}{i} \\ $$$$\Rightarrow{D}=\left(\mathrm{1},\mathrm{3}\right)+\left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}}{\mathrm{2}}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right) \\ $$$${B}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}−{isin}\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$$\Rightarrow{B}=\left(\mathrm{1},\mathrm{3}\right)+\left(\frac{\mathrm{9}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(\frac{\mathrm{11}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right) \\ $$

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