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Diberikan jajargenjang OABC dengan O ( 0 , 0 ) , A ( 2 , 0 ) , B ( 4 , 4 ) , dan C ( 2 , 4 ) . Tuliskan bayangan dari data berikut! b. Garis l ≡ garis BC dirotasi oleh [ A ( 2 , 0 ) , R ( − 6 0 ∘ ] .

Diberikan jajargenjang  dengan , dan . Tuliskan bayangan dari data berikut!

b. Garis  dirotasi oleh .

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O. Rahmawati

Master Teacher

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Jawaban terverifikasi

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Pembahasan

Menentukan garis dengan titik dan . Persamaan garis adalah Jadi, garis . Di rotasi oleh Kita akan melakukan dengan cara eliminasi Gauss-Jordan dalam menjawab soal ini. Persamaan matriks Gauss-Jordan: Proses Eliminasi Gauss-Jordan: Hal ini berarti: Kedua persamaan di atas disubstitusikan ke persamaan garis : Jadi, persamaan bayangan garis adalah .

  • Menentukan garis l identical to garis space B C dengan  titik straight B left parenthesis 4 comma 4 right parenthesis dan straight C left parenthesis 2 comma 4 right parenthesis.

          Persamaan garis B C adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses y minus y subscript 1 close parentheses open parentheses x subscript 2 minus x subscript 1 close parentheses end cell equals cell open parentheses x minus x subscript 1 close parentheses open parentheses y subscript 2 minus y subscript 1 close parentheses end cell row cell open parentheses y minus 4 close parentheses open parentheses 2 minus 4 close parentheses end cell equals cell open parentheses x minus 4 close parentheses open parentheses 4 minus 4 close parentheses end cell row cell left parenthesis y minus 4 right parenthesis left parenthesis negative 2 right parenthesis end cell equals cell open parentheses x minus 4 close parentheses open parentheses 0 close parentheses end cell row cell negative 2 y plus 8 end cell equals 0 row cell negative 2 y end cell equals cell negative 8 end cell row y equals 4 end table

Jadi, garis B C identical to l equals y minus 4 equals 0.

  • Di rotasi oleh open square brackets straight A left parenthesis 2 comma 0 right parenthesis comma space straight R left parenthesis negative 60 degree close square brackets

Kita akan melakukan dengan cara eliminasi Gauss-Jordan dalam menjawab soal ini.

Persamaan matriks Gauss-Jordan:

open parentheses table row cell cos space open parentheses negative 60 degree close parentheses end cell cell negative sin space open parentheses negative 60 degree close parentheses end cell row cell sin space open parentheses negative 60 degree close parentheses end cell cell cos space open parentheses negative 60 degree close parentheses end cell end table vertical ellipsis table row cell x apostrophe minus 2 end cell row cell y apostrophe minus 0 end cell end table close parentheses equals open parentheses table row cell 1 half end cell cell fraction numerator square root of 3 over denominator 2 end fraction end cell row cell negative fraction numerator square root of 3 over denominator 2 end fraction end cell cell 1 half end cell end table vertical ellipsis table row cell x apostrophe minus 2 end cell row cell y apostrophe end cell end table close parentheses

Proses Eliminasi Gauss-Jordan:

open parentheses table row cell 1 half end cell cell fraction numerator square root of 3 over denominator 2 end fraction end cell row cell negative fraction numerator square root of 3 over denominator 2 end fraction end cell cell 1 half end cell end table vertical ellipsis table row cell x apostrophe minus 2 end cell row cell y apostrophe end cell end table close parentheses table row cell 2 B subscript 1 end cell row cell 2 B subscript 2 end cell end table left right double arrow open parentheses table row 1 cell square root of 3 end cell row cell negative square root of 3 end cell 1 end table vertical ellipsis table row cell 2 x apostrophe minus 4 end cell row cell 2 y apostrophe end cell end table close parentheses table row blank row cell square root of 3 B subscript 1 plus B subscript 2 end cell end table left right double arrow open parentheses table row 1 cell square root of 3 end cell row 0 4 end table vertical ellipsis table row cell 2 x apostrophe minus 4 end cell row cell 2 square root of 2 x apostrophe minus 4 square root of 3 plus 2 y apostrophe end cell end table close parentheses table row blank row cell 1 fourth end cell end table left right double arrow open parentheses table row 1 cell square root of 3 end cell row 0 1 end table vertical ellipsis table row cell 2 x apostrophe minus 4 end cell row cell 1 half square root of 2 x apostrophe minus square root of 3 plus 1 half y apostrophe end cell end table close parentheses table row cell negative square root of 3 B subscript 2 plus B subscript 1 end cell row blank end table left right double arrow open parentheses table row 1 0 row 0 1 end table vertical ellipsis table row cell open parentheses 2 minus fraction numerator square root of 6 over denominator 2 end fraction close parentheses x apostrophe minus fraction numerator square root of 3 over denominator 2 end fraction y apostrophe minus 1 end cell row cell 1 half square root of 2 x apostrophe minus square root of 3 plus 1 half y apostrophe end cell end table close parentheses

Hal ini berarti:

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell equals cell open parentheses 2 minus fraction numerator square root of 6 over denominator 2 end fraction close parentheses x apostrophe minus fraction numerator square root of 3 over denominator 2 end fraction y apostrophe minus 1 end cell row x equals cell open parentheses 2 minus fraction numerator square root of 6 over denominator 2 end fraction close parentheses x apostrophe minus fraction numerator square root of 3 over denominator 2 end fraction y apostrophe plus 1 space... left parenthesis 1 right parenthesis end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell y minus 0 end cell equals cell 1 half square root of 2 x apostrophe minus square root of 3 plus 1 half y apostrophe end cell row y equals cell 1 half square root of 2 x apostrophe minus square root of 3 plus 1 half y apostrophe space... left parenthesis 2 right parenthesis end cell end table

Kedua persamaan di atas disubstitusikan ke persamaan garis y minus 4 equals 0:

table attributes columnalign right center left columnspacing 0px end attributes row cell y minus 4 end cell equals 0 row cell 1 half square root of 2 x apostrophe minus square root of 3 plus 1 half y apostrophe minus 4 end cell equals 0 row cell square root of 2 x apostrophe plus y apostrophe minus 2 square root of 3 minus 8 end cell equals 0 end table

Jadi, persamaan bayangan garis l identical to B C adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank y end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 3 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 8 end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table.

 

 

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