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Tentukan bayangan dari titik C(-2,1) oleh rotasi sebesar 90° dengan pusat O (0,0)!

Pertanyaan

Tentukan bayangan dari titik C(-2,1) oleh rotasi sebesar 90° dengan pusat O (0,0)!

Pembahasan Soal:

Hasil rotasi titik C terhadap pusat O sebesar 90° 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 0 cell negative 1 end cell row 1 0 end table close parentheses times open parentheses table row cell negative 2 end cell row 1 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 negative 1 end cell row cell negative 2 end cell end table close parentheses end style

Jadi, bayangan titik straight C adalah straight C apostrophe left parenthesis negative 1 comma negative 2 right parenthesis.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Damanhuri

Terakhir diupdate 11 Juli 2021

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Koordinat bayangan titik  setelah ditranslasi  dan dilanjutkan dengan rotasi berpusat di  sejauh  adalah ...

Pembahasan Soal:

Ingat konsep translasi titik begin mathsize 14px style open parentheses x comma space y close parentheses end style sejauh begin mathsize 14px style straight T equals open parentheses table row a row b end table close parentheses end style dan rotasi berpusat di begin mathsize 14px style straight O open parentheses 0 comma space 0 close parentheses end style sejauh begin mathsize 14px style 180 degree end style:

begin mathsize 14px style table row cell straight P open parentheses x comma space y close parentheses end cell cell rightwards arrow with T open parentheses table row a row b end table close parentheses on top end cell cell straight P apostrophe 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 x plus a end cell row cell y plus b end cell end table close parentheses end cell row cell straight P open parentheses x comma space y close parentheses end cell cell rightwards arrow with open square brackets R open parentheses 0 comma 0 close parentheses comma space 180 degree close square brackets on top end cell cell straight P apostrophe 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 negative 1 end cell 0 row 0 cell negative 1 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell end table end style  

 Koordinat bayangan titik begin mathsize 14px style straight P open parentheses 8 comma space 4 close parentheses end style ditranslasi sejauh begin mathsize 14px style open parentheses table row 3 row 1 end table close parentheses end style adalah

begin mathsize 14px style 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 8 row 4 end table close parentheses plus open parentheses table row 3 row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 8 plus 3 end cell row cell 4 plus 1 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 11 row 5 end table close parentheses end cell end table end style 

Diperoleh koordinat hasil translasi begin mathsize 14px style straight P apostrophe open parentheses 11 comma space 5 close parentheses end style, kemudian dilanjutkan rotasi berpusat di begin mathsize 14px style straight O open parentheses 0 comma space 0 close parentheses end style sejauh begin mathsize 14px style 180 degree end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe apostrophe end cell row cell y apostrophe apostrophe end cell end table close parentheses end cell 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 11 row 5 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 11 end cell row cell negative 5 end cell end table close parentheses end cell end table end style 

Dengan demikian bayangan titik P adalah begin mathsize 14px style P apostrophe apostrophe open parentheses negative 11 comma space minus 5 close parentheses end style

Oleh karena itu jawaban yang benar adalah B.

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Roboguru

Jika vektor dicerminkan pada garis x = y kemudian dirotasikan sejauh 90o dengan pusat (0,0) menjadi vektor v, maka u+v =

Pembahasan Soal:

straight u equals open parentheses table row straight a row straight b end table close parentheses rightwards arrow for straight M subscript straight x equals straight y end subscript of open parentheses table row straight b row straight a end table close parentheses rightwards arrow for straight R open square brackets open parentheses 0 comma 0 close parentheses comma 90 degree close square brackets of box enclose straight v equals open parentheses table row cell negative straight a end cell row straight b end table close parentheses end enclose  Sehingga  straight u plus straight v equals open parentheses table row straight a row straight b end table close parentheses plus open parentheses table row cell negative straight a end cell row straight b end table close parentheses equals open parentheses table row 0 row cell 2 straight b end cell end table close parentheses

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Roboguru

Diketahui segi empat dengan titik sudut , , , dan . Gambarlah pada kertas berpetak bangun hasil segi empat  pada masing-masing transformasi di bawah ini. Tuliskan koordinat titik hasil , , , dan  pada...

Pembahasan Soal:

  • Menentukan koordinat titik hasil perputaran pada titik O, sejauh 90 degree berlawana jarum jam. 

open parentheses table row cell x subscript o apostrophe end cell cell x subscript A apostrophe end cell cell x subscript B apostrophe end cell cell x subscript C apostrophe end cell row cell y subscript o apostrophe end cell cell y subscript A apostrophe end cell cell y subscript B apostrophe end cell cell y subscript C apostrophe end cell end table close parentheses equals open parentheses table row cell cos open parentheses 90 degree close parentheses end cell cell negative sin space degree left parenthesis 90 degree right parenthesis end cell row cell sin space left parenthesis 90 degree right parenthesis end cell cell cos space left parenthesis 90 degree right parenthesis end cell end table close parentheses open parentheses table row cell x subscript o end cell cell x subscript A end cell cell x subscript B end cell cell x subscript C end cell row cell y subscript o end cell cell y subscript A end cell cell y subscript B end cell cell y subscript C end cell end table close parentheses open parentheses table row cell x subscript o apostrophe end cell cell x subscript A apostrophe end cell cell x subscript B apostrophe end cell cell x subscript C apostrophe end cell row cell y subscript o apostrophe end cell cell y subscript A apostrophe end cell cell y subscript B apostrophe end cell cell y subscript C apostrophe end cell end table close parentheses equals open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row cell x subscript o end cell cell x subscript A end cell cell x subscript B end cell cell x subscript C end cell row cell y subscript o end cell cell y subscript A end cell cell y subscript B end cell cell y subscript C end cell end table close parentheses open parentheses table row cell x subscript o apostrophe end cell cell x subscript A apostrophe end cell cell x subscript B apostrophe end cell cell x subscript C apostrophe end cell row cell y subscript o apostrophe end cell cell y subscript A apostrophe end cell cell y subscript B apostrophe end cell cell y subscript C apostrophe end cell end table close parentheses equals open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row 0 4 0 4 row 0 0 5 5 end table close parentheses open parentheses table row cell x subscript o apostrophe end cell cell x subscript A apostrophe end cell cell x subscript B apostrophe end cell cell x subscript C apostrophe end cell row cell y subscript o apostrophe end cell cell y subscript A apostrophe end cell cell y subscript B apostrophe end cell cell y subscript C apostrophe end cell end table close parentheses equals open parentheses table row 0 0 cell negative 5 end cell cell negative 5 end cell row 0 4 0 4 end table close parentheses

Jadi, koordinat titik hasilnya adalah O apostrophe left parenthesis 0 comma 0 right parenthesis comma space A apostrophe left parenthesis 0 comma space 4 right parenthesis comma space B apostrophe left parenthesis negative 5 comma space 0 right parenthesis comma space C apostrophe left parenthesis negative 5 comma space 4 right parenthesis.

  • Berikut gambar hasil transformasinya:

0

Roboguru

Tentukan bayangan titik  setelah di rotasi dengan titik pusat  sejauh  berlawanan arah jarum jam !

Pembahasan Soal:

Matriks transformasi rotasi dengan titik pusat begin mathsize 14px style O open parentheses 0 comma 0 close parentheses end style sejauh begin mathsize 14px style 90 degree end style berlawanan dengan jarum jam adalah 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row T equals cell open parentheses table row cell cos open parentheses 90 degree close parentheses end cell cell negative sin open parentheses 90 degree close parentheses end cell row cell sin open parentheses 90 degree close parentheses end cell cell cos open parentheses 90 degree close parentheses end cell 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 end cell end table end style

Sehingga diperoleh bayangan titik begin mathsize 14px style B open parentheses negative 1 comma 9 close parentheses end style adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell B apostrophe end cell equals cell T times B end cell row blank equals cell open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses times open parentheses table row cell negative 1 end cell row 9 end table close parentheses end cell row blank equals cell open parentheses table row cell 0 times open parentheses negative 1 close parentheses plus open parentheses negative 1 close parentheses times 9 end cell row cell 1 times open parentheses negative 1 close parentheses plus 0 times 9 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 0 minus 9 end cell row cell negative 1 plus 0 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 9 end cell row cell negative 1 end cell end table close parentheses end cell end table end style

Dengan demikian bayangan titik begin mathsize 14px style straight B end style adalah begin mathsize 14px style B apostrophe open parentheses negative 9 comma negative 1 close parentheses end style.

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.

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