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Lingkaran ( x + 1 ) 2 + ( y − 2 ) 2 = 1 ditranslasikan oleh T 1 ​ = ( − 2 3 ​ ) dilanjutkan translasioleh T 2 ​ = ( 1 − 1 ​ ) .Hasil translasi lingkarantersebutadalah....

Lingkaran  ditranslasikan oleh  dilanjutkan translasi oleh . Hasil translasi lingkaran tersebut adalah ....

  1. begin mathsize 14px style left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 4 right parenthesis squared equals 1 end style  

  2. begin mathsize 14px style left parenthesis x plus 2 right parenthesis squared plus left parenthesis y plus 4 right parenthesis squared equals 1 end style   

  3. begin mathsize 14px style left parenthesis x minus 2 right parenthesis squared plus left parenthesis y minus 4 right parenthesis squared equals 1 end style  

  4. begin mathsize 14px style left parenthesis x plus 4 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals 1 end style 

  5. begin mathsize 14px style left parenthesis x plus 4 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 1 end style 

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M. Mariyam

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

jawaban yang tepat adalah A.

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Pembahasan

1) ditranslasikan oleh 2) dilanjutkan translasioleh Subsitusikan ke persamaan lingkaran : Jadi jawaban yang tepat adalah A.

1) ditranslasikan oleh begin mathsize 14px style T subscript 1 equals open parentheses table row cell negative 2 end cell row 3 end table close parentheses end style 

    size 14px x size 14px apostrophe size 14px equals size 14px x size 14px minus size 14px 2 size 14px space size 14px y size 14px apostrophe size 14px equals size 14px y size 14px plus size 14px 3 

2) dilanjutkan translasi oleh begin mathsize 14px style T subscript 2 equals open parentheses table row 1 row cell negative 1 end cell end table close parentheses end style

    size 14px x size 14px apostrophe size 14px apostrophe size 14px equals size 14px x size 14px apostrophe size 14px plus size 14px 1 size 14px space size 14px space size 14px space size 14px space size 14px equals size 14px x size 14px minus size 14px 2 size 14px plus size 14px 1 size 14px space size 14px space size 14px space size 14px space size 14px equals size 14px x size 14px minus size 14px 1 size 14px space size 14px space size 14px x size 14px equals size 14px x size 14px apostrophe size 14px apostrophe size 14px plus size 14px 1  size 14px y size 14px apostrophe size 14px apostrophe size 14px equals size 14px y size 14px apostrophe size 14px minus size 14px 1 size 14px space size 14px space size 14px space size 14px space size 14px equals size 14px y size 14px plus size 14px 3 size 14px minus size 14px 1 size 14px space size 14px space size 14px space size 14px space size 14px equals size 14px y size 14px plus size 14px 2 size 14px space size 14px space size 14px y size 14px equals size 14px y size 14px apostrophe size 14px apostrophe size 14px minus size 14px 2 

Subsitusikan ke persamaan lingkaran :

begin mathsize 14px style space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis x plus 1 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x squared plus 2 x plus 1 plus y squared minus 4 y plus 4 equals 1 space space space space space space space left parenthesis x plus 1 right parenthesis squared plus 2 open parentheses x plus 1 close parentheses plus 1 plus open parentheses y minus 2 close parentheses squared minus 4 open parentheses y minus 2 close parentheses plus 4 equals 1 x squared plus 2 x plus 1 plus 2 x plus 2 plus 1 plus y squared minus 4 y plus 4 minus 4 y plus 8 plus 4 equals 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x squared plus 4 x plus 4 plus y squared minus 8 y plus 16 equals 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses x plus 2 close parentheses squared plus open parentheses y minus 4 close parentheses squared equals 1 space end style  

Jadi jawaban yang tepat adalah A.

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Renita Sari

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