Roboguru

Diketahui segitiga  dengan titik sudut masing-masing ,  dan . Tentukan: a. Titik koordinat bayangan segitiga  pada rotasi  searah jarum jam dengan pusat rotasi titik asal .

Pertanyaan

Diketahui segitiga ABC dengan titik sudut masing-masing straight A open parentheses 2 comma space 1 close parenthesesstraight B open parentheses 4 comma space 2 close parentheses dan straight C open parentheses 2 comma space 4 close parentheses. Tentukan:

a. Titik koordinat bayangan segitiga ABC pada rotasi 90 degree searah jarum jam dengan pusat rotasi titik asal straight O open parentheses 0 comma space 0 close parentheses.

Pembahasan Soal:

Rotasi terhadap titik asal straight O open parentheses 0 comma space 0 close parentheses, dengan sudut rotasi theta, dapat dihitung dengan rumus 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 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

Diketahui segitiga ABC dengan titik sudut masing-masing straight A open parentheses 2 comma space 1 close parenthesesstraight B open parentheses 4 comma space 2 close parentheses dan straight C open parentheses 2 comma space 4 close parentheses.

Akan ditentukan bayangan segitiga ABC pada rotasi 90 degree searah jarum jam dengan pusat rotasi titik asal straight O open parentheses 0 comma space 0 close parentheses. Karena rotasi searah jarum jam, maka Rotasi bernilai negatif.

Menentukan bayangan titik straight A open parentheses 2 comma space 1 close parentheses.

straight A open parentheses 2 comma space 1 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses negative 90 degree close parentheses close square brackets on top straight A apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses 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 open parentheses negative 90 degree close parentheses end cell cell negative sin space open parentheses negative 90 degree close parentheses end cell row cell sin space open parentheses negative 90 degree close parentheses end cell cell cos space open parentheses negative 90 degree close parentheses end cell end table close parentheses open parentheses table row 2 row 1 end table close parentheses end cell row blank equals cell open parentheses table row 0 1 row cell negative 1 end cell 0 end table close parentheses open parentheses table row 2 row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 0 plus 1 end cell row cell negative 2 plus 0 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 1 row cell negative 2 end cell end table close parentheses end cell end table straight A open parentheses 2 comma space 1 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses negative 90 degree close parentheses close square brackets on top straight A apostrophe open parentheses 1 comma space minus 2 close parentheses

Menentukan bayangan titik straight B open parentheses 4 comma space 2 close parentheses.

straight B open parentheses 4 comma space 2 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses negative 90 degree close parentheses close square brackets on top straight B apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses 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 open parentheses negative 90 degree close parentheses end cell cell negative sin space open parentheses negative 90 degree close parentheses end cell row cell sin space open parentheses negative 90 degree close parentheses end cell cell cos space open parentheses negative 90 degree close parentheses end cell end table close parentheses open parentheses table row 4 row 2 end table close parentheses end cell row blank equals cell open parentheses table row 0 1 row cell negative 1 end cell 0 end table close parentheses open parentheses table row 4 row 2 end table close parentheses end cell row blank equals cell open parentheses table row cell 0 plus 2 end cell row cell negative 4 plus 0 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 2 row cell negative 4 end cell end table close parentheses end cell end table straight B open parentheses 4 comma space 2 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses negative 90 degree close parentheses close square brackets on top straight B apostrophe open parentheses 2 comma space minus 4 close parentheses

Menentukan bayangan titik straight C open parentheses 2 comma space 4 close parentheses.

straight C open parentheses 2 comma space 4 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses negative 90 degree close parentheses close square brackets on top straight C apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses 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 open parentheses negative 90 degree close parentheses end cell cell negative sin space open parentheses negative 90 degree close parentheses end cell row cell sin space open parentheses negative 90 degree close parentheses end cell cell cos space open parentheses negative 90 degree close parentheses 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 1 row cell negative 1 end cell 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 plus 4 end cell row cell negative 2 plus 0 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 4 row cell negative 2 end cell end table close parentheses end cell end table straight C open parentheses 2 comma space 4 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses negative 90 degree close parentheses close square brackets on top straight C apostrophe open parentheses 4 comma space minus 2 close parentheses

Jadi, diperoleh titik koordinat bayangan segitiga ABC adalah straight A apostrophe open parentheses 1 comma space minus 2 close parenthesesstraight B apostrophe open parentheses 2 comma space minus 4 close parentheses dan straight C apostrophe open parentheses 4 comma space minus 2 close parentheses

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Fathoni

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

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.

0

Roboguru

Garis  dirotasi oleh . Persamaan bayangan garis  yang terjadi berbentuk .... (Modifikasi UN 2014)

Pembahasan Soal:

Diketahui,

  • straight l identical to 3 x minus y minus 1 equals 0
  • Dirotasikan 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 4 close parentheses close square brackets end style 

Ditanyakan,

  • Bayangan garis

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 45 end cell cell negative sin space 45 end cell row cell sin space 45 end cell cell cos space 45 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 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 open parentheses table row x row y 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 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 end cell row blank blank blank row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator open parentheses begin display style 1 half square root of 2 space end root times 1 half square root of 2 end style close parentheses minus open parentheses begin display style 1 half square root of 2 times negative 1 half square root of 2 end style 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 end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator begin display style 2 over 4 minus open parentheses negative 2 over 4 close parentheses 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 end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator begin display style 1 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 end cell row cell open parentheses table row x row y 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 x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell 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 end cell end table

Maka di dapat x equals 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe dan y equals negative 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe

Masukan ke persamaan garis straight l identical to 3 x minus y minus 1 equals 0,

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x minus y minus 1 end cell equals 0 row cell 3 open parentheses 1 half square root of 2 x apostrophe plus 1 half square root of 2 y apostrophe close parentheses minus open parentheses negative 1 half x apostrophe plus 1 half square root of 2 y apostrophe close parentheses minus 1 end cell equals 0 row cell 3 over 2 square root of 2 x apostrophe plus 3 over 2 square root of 2 y apostrophe plus 1 half x apostrophe minus 1 half square root of 2 y apostrophe minus 1 end cell equals 0 row cell 4 over 2 square root of 2 x apostrophe plus 2 over 2 square root of 2 y apostrophe minus 1 end cell equals 0 row cell 2 square root of 2 x apostrophe plus square root of 2 y apostrophe minus 1 end cell equals 0 row cell 4 x apostrophe plus 2 y apostrophe minus square root of 2 end cell equals 0 row blank blank blank end table

Jadi, bayangan garisnya adalah .table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 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 apostrophe 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 2 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 apostrophe 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 cell square root of 2 end cell 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

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Pada gambar di bawah ini tuliskan hasil bangun perputaran terhadap titik  pada: d. sumbu  dengan putaran sejauh  searah jarum jam.

Pembahasan Soal:

Jadi, sumbu undefined dengan putaran sejauh begin mathsize 14px style 300 degree end style searah jarum jam di gambarkan seperti di atas.

0

Roboguru

Persamaan peta suatu kurva oleh rotasi pusat O bersudut , dilanjutkan dilatasi  adalah . Persamaan kurva semula adalah ....

Pembahasan Soal:

Persamaan transformasi dengan pusat straight O left parenthesis 0 comma space 0 right parenthesis dan sudut rotasi theta berlawanan arah jarum jam, ditulis straight R left square bracket 0 comma space theta right square bracket, untuk pemetaan left parenthesis x comma space y right parenthesis ke left parenthesis x apostrophe comma space y apostrophe right parenthesis dapat dinyatakan sebagai berikut.

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 theta end cell cell negative sin theta end cell row cell sin theta end cell cell cos theta end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell end table   

Transformasi dilatasi terhadap pusat straight O open parentheses 0 comma space 0 close parentheses dengan faktor skala k, ditulis left square bracket straight O comma space k right square bracket untuk pemetaan dari left parenthesis x comma space y right parenthesis ke left parenthesis x apostrophe comma space y apostrophe right parenthesis, dapat dinyatakan sebagai berikut.

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals k open parentheses table row x row y end table close parentheses

Berdasarkan aturan diatas, sehingga rotasi straight R open square brackets 0 comma space straight pi over 2 close square brackets adalah sebagai berikut.

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 90 degree end cell cell negative sin 90 degree end cell row cell sin 90 degree end cell cell cos 90 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 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 0 times x plus open parentheses negative 1 close parentheses times y end cell row cell 1 times x plus 0 times y 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 y end cell row x end table close parentheses end cell end table             

Kemudian, dilanjutkan dilatasi open square brackets 0 comma space 2 close square brackets adalah sebagai berikut.

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 2 open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell 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 2 open parentheses table row cell negative y end cell row x end table close parentheses end cell 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 2 y end cell row cell 2 x end cell end table close parentheses end cell end table      

Berdasarkan kesamaan matriks diatas diperoleh:

x apostrophe apostrophe equals negative 2 y y apostrophe apostrophe equals 2 x     

Substitusikan nilai x apostrophe apostrophe dan y apostrophe apostrophe yang diperoleh ke persamaan peta.

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell 2 plus y minus y squared end cell row cell x apostrophe apostrophe end cell equals cell 2 plus y apostrophe apostrophe minus y apostrophe apostrophe squared end cell row cell negative 2 y end cell equals cell 2 plus 2 x minus open parentheses 2 x close parentheses squared end cell row cell fraction numerator negative 2 y over denominator negative 2 end fraction end cell equals cell fraction numerator 2 plus 2 x minus 4 x squared over denominator negative 2 end fraction end cell row y equals cell negative 1 minus x plus 2 x squared end cell row y equals cell 2 x squared minus x minus 1 end cell end table     

Dengan demikian persamaan kurva semula adalah y equals 2 x squared minus x minus 1.

 Jadi, jawaban yang tepat adalah E.

0

Roboguru

Bayangan titik  oleh rotasi  adalah ...

Pembahasan Soal:

Diketahui:

Titik begin mathsize 14px style P open parentheses 2 comma space minus sign 3 close parentheses end style.

Rotasi begin mathsize 14px style R open square brackets O comma space 90 degree close square brackets end style.

Rotasi adalah transformasi yang mengubah koordinat suatu titik terhadap titik tetap dengan besar tertentu dan arah tertentu. 

Bayangan rotasi sebesar begin mathsize 14px style 90 degree end style dengan pusat begin mathsize 14px style open parentheses 0 comma space 0 close parentheses end style, yaitu:

begin mathsize 14px style straight P open parentheses x comma space y close parentheses rightwards arrow straight P apostrophe open parentheses negative y comma space x close parentheses end style

Sehingga diperoleh:

begin mathsize 14px style table attributes columnalign left center left end attributes row cell straight P open parentheses x comma space y close parentheses end cell rightwards arrow cell straight P apostrophe open parentheses negative y comma space x close parentheses end cell row cell straight P open parentheses 2 comma space minus 3 close parentheses end cell rightwards arrow cell straight P apostrophe open parentheses negative open parentheses negative 3 close parentheses comma space 2 close parentheses end cell row cell straight P open parentheses 2 comma space minus 3 close parentheses end cell rightwards arrow cell straight P apostrophe open parentheses 3 comma space 2 close parentheses end cell end table end style

Bayangan titik begin mathsize 14px style P open parentheses 2 comma space minus sign 3 close parentheses end style dirotasi dengan begin mathsize 14px style R open square brackets O comma space 90 degree close square brackets end style adalah begin mathsize 14px style straight P to the power of apostrophe open parentheses 3 comma space 2 close parentheses end style.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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