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Lingkaran L:(x−2)2+(y−2)2=1 direfleksikan terhadap sumbu X, lalu ditranslasikan oleh T=[−13​]. Persamaan lingkaran hasil transformasi adalah ...

Pertanyaan

Lingkaran begin mathsize 14px style L colon left parenthesis x minus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 1 end style direfleksikan terhadap sumbu begin mathsize 14px style X end style, lalu ditranslasikan oleh begin mathsize 14px style T equals open square brackets table row cell negative 1 end cell row 3 end table close square brackets end style. Persamaan lingkaran hasil transformasi adalah undefined 

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

  2. size 14px left parenthesis size 14px x size 14px minus size 14px 1 size 14px right parenthesis to the power of size 14px 2 size 14px plus size 14px left parenthesis size 14px y size 14px minus size 14px 1 size 14px right parenthesis to the power of size 14px 2 size 14px equals size 14px 1 

  3. size 14px left parenthesis size 14px x size 14px minus size 14px 1 size 14px right parenthesis to the power of size 14px 2 size 14px plus size 14px left parenthesis size 14px y size 14px plus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px equals size 14px 1 

  4. size 14px left parenthesis size 14px x size 14px minus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px plus size 14px left parenthesis size 14px y size 14px plus size 14px 1 size 14px right parenthesis to the power of size 14px 2 size 14px equals size 14px 1 

  5. size 14px left parenthesis size 14px x size 14px plus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px plus size 14px left parenthesis size 14px y size 14px plus size 14px 1 size 14px right parenthesis to the power of size 14px 2 size 14px equals size 14px 1 

H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B  

Pembahasan

Misalkan titik begin mathsize 14px style open parentheses x comma space y close parentheses end style terletak pada lingkaran begin mathsize 14px style L end style.

I) Hasil refleksii terhadap sumbu begin mathsize 14px style x end style 

begin mathsize 14px style open square brackets table row cell x apostrophe end cell row cell y apostrophe end cell end table close square brackets equals open square brackets table row 1 0 row 0 cell negative 1 end cell end table close square brackets open square brackets table row x row y end table close square brackets equals open square brackets table row x row cell negative y end cell end table close square brackets end style 

II) Hasil translasi begin mathsize 14px style T equals open square brackets table row cell negative 1 end cell row 3 end table close square brackets end style titik begin mathsize 14px style open parentheses x apostrophe comma space y apostrophe close parentheses end style

begin mathsize 14px style open square brackets table row cell x " end cell row cell y " end cell end table close square brackets equals open square brackets table row cell x apostrophe plus a end cell row cell y apostrophe plus b end cell end table close square brackets equals open square brackets table row cell x minus 1 end cell row cell negative y plus 3 end cell end table close square brackets end style 

Berdasarkan kesamaan matriks diperoleh 

begin mathsize 14px style x " equals x minus 1 left right arrow x equals x " plus 1 y " equals negative y plus 3 left right arrow y equals negative y " plus 3 end style 

Subtitusikan bentuk undefined dan begin mathsize 14px style y end style ke dalam persamaan lingkaran 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x minus 2 close parentheses squared plus open parentheses y minus 2 close parentheses squared end cell equals 1 row cell open parentheses x " plus 1 minus 2 close parentheses squared plus open parentheses negative y " plus 3 minus 2 close parentheses squared end cell equals 1 row cell open parentheses x " minus 1 close parentheses squared plus open parentheses negative y " plus 1 close parentheses squared end cell equals 1 row cell open parentheses x " minus 1 close parentheses squared plus open parentheses y " minus 1 close parentheses squared end cell equals 1 end table end style 

Jadi, jawaban yang tepat adalah B  

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