Roboguru

Tentukan bayangan titik A(6,3) akibat diputar dengan aturan sebagai berikut: c. 180∘ dengan pusat O(0,0).

Pertanyaan

Tentukan bayangan titik begin mathsize 14px style straight A open parentheses 6 comma space 3 close parentheses end style akibat diputar dengan aturan sebagai berikut:

c. begin mathsize 14px style 180 degree end style dengan pusat undefined.

Pembahasan Soal:

Untuk menentukan bayangan rotasi dengan pusat begin mathsize 14px style straight O left parenthesis 0 comma space 0 right parenthesis end style sejauh begin mathsize 14px style theta end style adalah sebagai berikut:

begin mathsize 14px style open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell cos space theta end cell cell negative sin space theta end cell row cell sin space theta end cell cell cos space theta end cell end table close parentheses open parentheses table row x row y end table close parentheses end style 

Diketahui:
Titik begin mathsize 14px style straight A open parentheses 6 comma space 3 close parentheses end style diputar atau dirotasi begin mathsize 14px style 180 degree end style dengan pusat undefined.
Akan ditentukan bayangan titik tersebut yaitu begin mathsize 14px style straight A apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses end style.

Perhatikan perhitungan berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space 180 degree end cell cell negative sin space 180 degree end cell row cell sin space 180 degree end cell cell cos space 180 degree end cell end table close parentheses open parentheses table row 6 row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table close parentheses open parentheses table row 6 row 3 end table close parentheses end cell row blank equals cell open parentheses table row cell open parentheses negative 1 close parentheses cross times 6 plus 0 cross times 3 end cell row cell 0 cross times 6 plus open parentheses negative 1 close parentheses cross times 3 end cell end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell negative 6 end cell row cell negative 3 end cell end table close parentheses end cell end table end style  

Jadi, diperoleh bayangannya adalah begin mathsize 14px style straight A apostrophe open parentheses negative 6 comma space minus 3 close parentheses end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Dwi

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 05 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Kurva xy=6 dirotasi oleh [O(0,0),R(2π​)] menghasilkan bayangan kurva dengan persamaan .... (OCR 2012)

Pembahasan Soal:

Diketahui,

  • begin mathsize 14px style x y equals 6 end style 
  • Dirotasi oleh begin mathsize 14px style open square brackets straight O open parentheses 0 comma space 0 close parentheses comma space straight R open parentheses straight pi over 2 close parentheses close square brackets end style

Ditanyakan,

  • Bayangan kurva

Gunakan persamaan matriks berikut.

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell cos space straight theta end cell cell negative sin space straight theta end cell row cell sin space straight theta end cell cell cos space straight theta end cell end table close parentheses open parentheses table row x row y end table close parentheses

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space 90 end cell cell negative sin space 90 end cell row cell sin space 90 end cell cell cos space 90 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell negative y end cell row x end table close parentheses end cell row blank blank blank row cell x apostrophe end cell equals cell negative y end cell row cell y apostrophe end cell equals cell negative x apostrophe end cell end table

Kemudian subtitusikan ke begin mathsize 14px style x y equals 6 end style ,

table attributes columnalign right center left columnspacing 0px end attributes row cell x y end cell equals 6 row cell left parenthesis negative y apostrophe right parenthesis left parenthesis x right parenthesis end cell equals 6 row cell negative x apostrophe y apostrophe end cell equals 6 row cell x y end cell equals cell negative 6 end cell end table

Jadi, bayangan kurva adalah x y equals negative 6.


Oleh karena itu, jawaban yang benar adalah A. 

0

Roboguru

Bayangan titik S (2,4) oleh rotasi yang berpusat di O (0,0) sejauh 90o berlawanan arah jarum jam dan dilanjutkan oleh pencerminan terhadap garis y = x adalah...

Pembahasan Soal:

T subscript 1 equals open parentheses table row cell cos space 90 to the power of o end cell cell negative sin space 90 to the power of o end cell row cell sin space 90 to the power of o end cell cell cos space 90 to the power of o end cell end table close parentheses equals open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses  T subscript 1 equals open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses  M e n c a r i space b a y a n g a n space t i t i k space S  open parentheses table row cell X apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row 0 1 row 1 0 end table close parentheses open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row 2 row 4 end table close parentheses equals open parentheses table row 1 0 row 0 cell negative 1 end cell end table close parentheses open parentheses table row 2 row cell negative 4 end cell end table close parentheses

0

Roboguru

Tentukan bayangan garis≡y=2x+2​ jika dirotasi sejauh 45∘ dengan pusat (0,0).

Pembahasan Soal:

  • Gunakan persamaan matriks berikut:

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell cos space 45 degree end cell cell negative sin space 45 degree end cell row cell sin space 45 degree space end cell cell cos space 45 degree end cell end table close parentheses open parentheses table row x row y end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell 1 half square root of 2 end cell cell negative 1 half square root of 2 end cell row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row x row y end table close parentheses

  • Menentukan persamaan bayangan dengan cara invers matriks:

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell 1 half square root of 2 end cell cell negative 1 half square root of 2 end cell row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row x row y end table close parentheses open parentheses table row x row y end table close parentheses equals open parentheses table row cell 1 half square root of 2 end cell cell negative 1 half square root of 2 end cell row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses to the power of negative 1 end exponent open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses open parentheses table row x row y end table close parentheses equals fraction numerator 1 over denominator open parentheses 1 half square root of 2 close parentheses open parentheses 1 half square root of 2 close parentheses minus open parentheses negative 1 half square root of 2 close parentheses open parentheses 1 half square root of 2 close parentheses end fraction open parentheses table row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses open parentheses table row x row y end table close parentheses equals fraction numerator 1 over denominator begin display style 1 half plus 1 half end style end fraction open parentheses table row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses open parentheses table row x row y end table close parentheses equals open parentheses table row cell 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell row cell negative 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell end table close parentheses open parentheses table row x row y end table close parentheses equals 1 half square root of 2 open parentheses table row cell x apostrophe plus y apostrophe end cell row cell negative x apostrophe plus y apostrophe end cell end table close parentheses

diperoleh x equals 1 half square root of 2 open parentheses x apostrophe plus y apostrophe close parentheses space dan space y equals 1 half square root of 2 open parentheses negative x apostrophe plus y apostrophe close parentheses, substitusikan persamaan garis y equals 2 x plus square root of 2.

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 2 x plus square root of 2 end cell row cell 1 half square root of 2 left parenthesis negative x apostrophe plus y apostrophe right parenthesis end cell equals cell 2 open parentheses 1 half square root of 2 open parentheses x apostrophe plus y apostrophe close parentheses close parentheses plus square root of 2 end cell row cell negative 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe end cell equals cell square root of 2 x apostrophe plus square root of 2 y apostrophe plus square root of 2 end cell row cell negative 3 over 2 square root of 2 x apostrophe minus 1 half square root of 2 y apostrophe minus square root of 2 end cell equals 0 row cell 3 over 2 square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe plus square root of 2 end cell equals 0 end table

Jadi, bayangan garis adalah 3 over 2 square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe plus square root of 2 equals 0.

0

Roboguru

Titik A(x,y) dirotasi sebesar 45° berlawanan arah jarum jam dengan pusat O(0,0) menghasilkan bayangan . Bayangan dari titik A jika dirotasi sebesar 45° searah jarum jam dengan pusat O(0,0) adalah ....

Pembahasan Soal:

Ingat bahwa titik A(x, y) jika dirotasi dengan sudut rotasi sebesar θ berlawanan arah jarum jam dengan pusat O(0, 0), maka didapat titik bayangan A’(x’,y’) dengan

undefined

Pada soal diketahui titik A dirotasi sebesar 45° berlawanan arah jarum jam dengan pusat O(0,0) menghasilkan bayangan undefined
Sehingga θ = 45°, begin mathsize 14px style x to the power of apostrophe equals 2 square root of 2 space d a n space y to the power of apostrophe equals negative 6 square root of 2 end style.

Maka didapatkan hubungan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell 2 square root of 2 end cell row cell negative 6 square root of 2 end cell end table close parentheses end cell equals cell open parentheses table row cell cos invisible function application 45 degree end cell cell negative sin invisible function application 45 degree end cell row cell sin invisible function application 45 degree end cell cell cos invisible function application 45 degree end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell 2 square root of 2 end cell row cell negative 6 square root of 2 end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 end cell cell negative 1 half square root of 2 end cell row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell 1 half square root of 2 open parentheses table row 4 row cell negative 12 end cell end table close parentheses end cell equals cell 1 half square root of 2 open parentheses table row 1 cell negative 1 end cell row 1 1 end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row 4 row cell negative 12 end cell end table close parentheses end cell equals cell open parentheses table row 1 cell negative 1 end cell row 1 1 end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row 4 row cell negative 12 end cell end table close parentheses end cell equals cell open parentheses table row cell x minus y end cell row cell x plus y end cell end table close parentheses end cell end table end style

Sehingga didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x minus y end cell equals 4 row cell x plus y end cell equals cell negative 12 end cell end table end style

dengan menggunakan metode eliminasi, didapat bahwa

begin mathsize 14px style table row cell x minus y equals 4 end cell blank row cell bottom enclose x plus y equals negative 12 end enclose end cell plus row cell 2 x equals negative 8 end cell blank row cell x equals negative 4 end cell blank end table end style

Substitusi nilai x ke salah satu persamaan, misal persamaan kedua, maka didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x plus y end cell equals cell negative 12 end cell row cell negative 4 plus y end cell equals cell negative 12 end cell row y equals cell negative 12 plus 4 end cell row y equals cell negative 8 end cell end table end style

Sehingga didapat titik A(-4, -8).

Selanjutnya titik A(-4, -8) dirotasi sebesar 45° searah jarum jam dengan pusat O(0,0). Sehingga didapat x = -4, y = -8, dan θ = -45°.
Maka didapat hubungan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos invisible function application open parentheses negative 45 degree close parentheses end cell cell negative sin invisible function application open parentheses negative 45 degree close parentheses end cell row cell sin invisible function application open parentheses negative 45 degree close parentheses end cell cell cos invisible function application open parentheses negative 45 degree close parentheses end cell end table close parentheses open parentheses table row cell negative 4 end cell row cell negative 8 end cell end table close parentheses end cell row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 end cell cell negative open parentheses negative 1 half square root of 2 close parentheses end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell negative 4 end cell row cell negative 8 end cell end table close parentheses end cell row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 end cell cell 1 half square root of 2 end cell row cell negative 1 half square root of 2 end cell cell 1 half square root of 2 end cell end table close parentheses open parentheses table row cell negative 4 end cell row cell negative 8 end cell end table close parentheses end cell row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 2 open parentheses negative 4 close parentheses plus 1 half square root of 2 open parentheses negative 8 close parentheses end cell row cell negative 1 half square root of 2 open parentheses negative 4 close parentheses plus 1 half square root of 2 open parentheses negative 8 close parentheses end cell end table close parentheses end cell row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell negative 2 square root of 2 minus 4 square root of 2 end cell row cell 2 square root of 2 minus 4 square root of 2 end cell end table close parentheses end cell row cell open parentheses table row cell x to the power of apostrophe end cell row cell y to the power of apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell negative 6 square root of 2 end cell row cell negative 2 square root of 2 end cell end table close parentheses end cell end table end style

Sehingga didapat titik bayangannya adalah begin mathsize 14px style A subscript 2 superscript apostrophe open parentheses negative 6 square root of 2 comma negative 2 square root of 2 close parentheses end style.

0

Roboguru

Titik A(2,4) dirotasi sejauh 90∘ searah jarum jam terhadap pusat O(0,0). Koordinat bayangan titik A sama dengan .... (Kinomatika 2014)

Pembahasan Soal:

Diketahui,

  •  begin mathsize 14px style straight A left parenthesis 2 comma space 4 right parenthesis end style 
  • Rotasi sejauh begin mathsize 14px style 90 degree end style searah jarum jam
  • Terhadap pusat begin mathsize 14px style straight O left parenthesis 0 comma space 0 right parenthesis end style

Ditanyakan,

  • Koordinat bayangan titik begin mathsize 14px style straight A end style

Rumus mencari koordinat bayangan titik,

open parentheses table row cell x apostrophe subscript A end cell row cell y apostrophe subscript A end cell end table close parentheses equals open parentheses table row cell cos space straight theta end cell cell negative sin space straight theta end cell row cell sin space straight theta end cell cell cos space straight theta end cell end table close parentheses open parentheses table row cell x subscript A end cell row cell y subscript A end cell end table close parentheses

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe subscript A end cell row cell y apostrophe subscript A end cell end table close parentheses end cell equals cell open parentheses table row cell cos space 90 end cell cell negative sin space 90 end cell row cell sin space 90 end cell cell cos space 90 end cell end table close parentheses open parentheses table row 2 row 4 end table close parentheses end cell row blank equals cell open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row 2 row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell 0 times 2 plus left parenthesis negative 1 right parenthesis times 4 end cell row cell 1 times 2 plus 0 times 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 0 minus 4 end cell row cell 2 plus 0 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 end cell row 2 end table close parentheses end cell end table

Jadi, koordinat bayangan titik begin mathsize 14px style straight A end style adalah left parenthesis negative 4 comma 2 right parenthesis.


Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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