Roboguru

Diketahui vektor posisi , , dan . Titik P berada di AB dengan perbandingan , titik Q berada di antara AC dengan perbandingan , titik R merupakan perpotongan ruas garis BQ dan CP. Tentukan: a. koordinat titik P b. koordina titik Q c. koordinat titik R

Pertanyaan

Diketahui vektor posisi OA with rightwards arrow on top equals a with rightwards arrow on top equals open parentheses table row 1 row 7 end table close parentheses, OB with rightwards arrow on top equals b with rightwards arrow on top equals open parentheses table row cell negative 4 end cell row cell negative 3 end cell end table close parentheses, dan OC with rightwards arrow on top equals c with rightwards arrow on top equals open parentheses table row 5 row 3 end table close parentheses. Titik P berada di AB dengan perbandingan AP space colon space PB equals 3 space colon space 2, titik Q berada di antara AC dengan perbandingan text AQ : QC=1 : 3 end text, titik R merupakan perpotongan ruas garis BQ dan CP. Tentukan:

a. koordinat titik P

b. koordina titik Q

c. koordinat titik Rbegin mathsize 14px style space end style 

Pembahasan Soal:

Vektor posisi adalah vektor yang berpangkal di pusat koordinat open parentheses 0 comma space 0 close parentheses dan berujung di suatu titik open parentheses x comma space y close parentheses.

Vektor stack A B with rightwards arrow on top dapat diperoleh dari vektor posisi titik B dikurangi vektor posisi titik A

stack A B with rightwards arrow on top equals b with rightwards arrow on top minus a with rightwards arrow on top 

Permasalahan di atas dapat gambarkan sebagai berikut.



 

a. koordinat titik P

Diketahui perbandingan AP space colon space PB equals 3 space colon space 2.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator AP with rightwards arrow on top over denominator PB with rightwards arrow on top end fraction end cell equals cell 3 over 2 end cell row cell fraction numerator p with rightwards arrow on top minus a with rightwards arrow on top over denominator b with rightwards arrow on top minus p with rightwards arrow on top end fraction end cell equals cell 3 over 2 end cell row cell 2 open parentheses p with rightwards arrow on top minus a with rightwards arrow on top close parentheses end cell equals cell 3 open parentheses b with rightwards arrow on top minus p with rightwards arrow on top close parentheses end cell row cell 2 p with rightwards arrow on top minus 2 a with rightwards arrow on top end cell equals cell 3 b with rightwards arrow on top minus 3 p with rightwards arrow on top end cell row cell 5 p with rightwards arrow on top end cell equals cell 2 a with rightwards arrow on top plus 3 b with rightwards arrow on top end cell row cell p with rightwards arrow on top end cell equals cell 1 fifth times open parentheses 2 a with rightwards arrow on top plus 3 b with rightwards arrow on top close parentheses end cell row cell p with rightwards arrow on top end cell equals cell 1 fifth times open parentheses 2 open parentheses table row 1 row 7 end table close parentheses plus 3 open parentheses table row cell negative 4 end cell row cell negative 3 end cell end table close parentheses close parentheses end cell row cell p with rightwards arrow on top end cell equals cell 1 fifth times open parentheses table row cell 2 minus 12 end cell row cell 14 minus 9 end cell end table close parentheses end cell row cell p with rightwards arrow on top end cell equals cell 1 fifth times open parentheses table row cell negative 10 end cell row 5 end table close parentheses end cell row cell p with rightwards arrow on top end cell equals cell open parentheses table row cell negative 2 end cell row 1 end table close parentheses end cell row blank blank cell therefore space straight P open parentheses negative 2 comma space 1 close parentheses end cell end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight P end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 2 comma space 1 close parentheses end cell end table.

b. koordina titik Q

Diketahui perbandingan text AQ : QC=1 : 3 end text

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator AQ with rightwards arrow on top over denominator QC with rightwards arrow on top end fraction end cell equals cell 1 third end cell row cell fraction numerator q with rightwards arrow on top minus a with rightwards arrow on top over denominator c with rightwards arrow on top minus q with rightwards arrow on top end fraction end cell equals cell 1 third end cell row cell 3 open parentheses q with rightwards arrow on top minus a with rightwards arrow on top close parentheses end cell equals cell c with rightwards arrow on top minus q with rightwards arrow on top end cell row cell 4 q with rightwards arrow on top end cell equals cell 3 a with rightwards arrow on top plus c with rightwards arrow on top end cell row cell q with rightwards arrow on top end cell equals cell 1 fourth open parentheses 3 a with rightwards arrow on top plus c with rightwards arrow on top close parentheses end cell row cell q with rightwards arrow on top end cell equals cell 1 fourth times open parentheses 3 open parentheses table row 1 row 7 end table close parentheses plus open parentheses table row 5 row 3 end table close parentheses close parentheses end cell row cell q with rightwards arrow on top end cell equals cell 1 fourth times open parentheses table row cell 3 plus 5 end cell row cell 21 plus 3 end cell end table close parentheses end cell row cell q with rightwards arrow on top end cell equals cell 1 fourth times open parentheses table row 8 row 24 end table close parentheses end cell row cell q with rightwards arrow on top end cell equals cell open parentheses table row 2 row 6 end table close parentheses end cell row blank blank cell therefore space straight Q open parentheses 2 comma space 6 close parentheses end cell end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight Q end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 2 comma space 6 close parentheses end cell end table.

c. koordinat titik Rbegin mathsize 14px style space end style 

Tentukan persamaan garis CP melalui titik straight C open parentheses 5 comma space 3 close parentheses dan straight P open parentheses negative 2 comma space 1 close parentheses.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction end cell equals cell fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction end cell row cell fraction numerator y minus 3 over denominator 1 minus 3 end fraction end cell equals cell fraction numerator x minus 5 over denominator negative 2 minus 5 end fraction end cell row cell negative 7 open parentheses y minus 3 close parentheses end cell equals cell negative 2 open parentheses x minus 5 close parentheses end cell row cell negative 7 y plus 21 end cell equals cell negative 2 x plus 10 end cell row 11 equals cell negative 2 x plus 7 y end cell row cell negative 2 x plus 7 y end cell equals cell 11 space... open parentheses straight i close parentheses end cell end table 

Tentukan persmaan garis BQ melalui titik straight B open parentheses negative 4 comma space minus 3 close parentheses dan straight Q open parentheses 2 comma space 6 close parentheses.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction end cell equals cell fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction end cell row cell fraction numerator y plus 3 over denominator 6 plus 3 end fraction end cell equals cell fraction numerator x plus 4 over denominator 2 plus 4 end fraction end cell row cell 6 open parentheses y plus 3 close parentheses end cell equals cell 9 open parentheses x plus 4 close parentheses end cell row cell 6 y plus 18 end cell equals cell 9 x plus 36 end cell row cell negative 9 x plus 6 y end cell equals cell 18 space... open parentheses ii close parentheses end cell end table 

Titik R merupakan perpotongan ruas garis BQ dan CP.

Eliminasi persamaan (i) dan (ii) untuk menentukan koordinat titik R. 

negative 2 x plus 7 y equals 11 space open vertical bar times 9 close vertical bar space minus 18 x plus 63 y equals 99 minus 9 x plus 6 y equals 18 space open vertical bar times 2 close vertical bar space bottom enclose negative 18 x plus 12 y equals 36 end enclose space minus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 51 y equals 63 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y equals 63 over 51 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y equals 21 over 17 

Substitusikan y equals 21 over 17 pada persamaan (i).

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 2 x plus 7 y end cell equals 11 row cell negative 2 x end cell equals cell 11 minus 7 y end cell row cell negative 2 x end cell equals cell 11 minus 7 times 21 over 17 end cell row cell negative 2 x end cell equals cell fraction numerator 187 minus 147 over denominator 17 end fraction end cell row x equals cell 40 over 17 times fraction numerator 1 over denominator negative 2 end fraction end cell row x equals cell negative 20 over 17 end cell end table 

Sehingga, straight R open parentheses negative 20 over 17 comma space 21 over 17 close parentheses.

Jadi, straight P open parentheses negative 2 comma space 1 close parenthesesstraight Q open parentheses 2 comma space 6 close parentheses, dan straight R open parentheses negative 20 over 17 comma space 21 over 17 close parentheses.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Indraprasta PGRI

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

1. Diketahui titik .  Tentukan vektor: a.

Pembahasan Soal:

Vektor tersebut

undefined 

Jadi, vektor stack A B with rightwards arrow on top space equals space open parentheses table row 1 row cell negative 8 end cell end table close parentheses  

0

Roboguru

Pada , E titik tengah QR dan D adalah titik berat segitiga tersebut. Jika  dan , maka ruas garis berarah DE dapat dinyatakan sebagai ...

Pembahasan Soal:

Garis berat sebuah segitiga adalah garis yang melalui sebuah titik sudut dan membagi sisi di depan sudut menjadi dua bagian sama panjang.

Titik berat segitiga adalah titik perpotongan antara ketiga garis berat segitiga.

Diketahui E titik tengah QR dan D adalah titik berat increment PQR, maka:

QE equals ER equals 1 half QR

serta PE adalah garis berat increment PQR, akibatnya:

PD over DE equals 2 over 1 

Jika a with rightwards arrow on top equals PQ with rightwards arrow on top dan b with rightwards arrow on top equals PR with rightwards arrow on top, maka:

QR with rightwards arrow on top equals QP with rightwards arrow on top plus PR with rightwards arrow on top QR with rightwards arrow on top equals negative a with rightwards arrow on top plus b with rightwards arrow on top 

Sehingga, PE with rightwards arrow on top dapat dinyatakan sebagai:

table attributes columnalign right center left columnspacing 0px end attributes row cell PE with rightwards arrow on top end cell equals cell PQ with rightwards arrow on top plus QE with rightwards arrow on top end cell row cell PE with rightwards arrow on top end cell equals cell PQ with rightwards arrow on top plus 1 half QR with rightwards arrow on top end cell row cell PE with rightwards arrow on top end cell equals cell a with rightwards arrow on top plus 1 half open parentheses negative a with rightwards arrow on top plus b with rightwards arrow on top close parentheses end cell row cell PE with rightwards arrow on top end cell equals cell a with rightwards arrow on top minus 1 half a with rightwards arrow on top plus 1 half b with rightwards arrow on top end cell row cell PE with rightwards arrow on top end cell equals cell 1 half a with rightwards arrow on top plus 1 half b with rightwards arrow on top end cell end table 

Maka, DE with rightwards arrow on top dapat dinyatakan sebagai:

DE with rightwards arrow on top equals 1 third PE with rightwards arrow on top DE with rightwards arrow on top equals 1 third open parentheses 1 half a with rightwards arrow on top plus 1 half b with rightwards arrow on top close parentheses DE with rightwards arrow on top equals 1 over 6 a with rightwards arrow on top plus 1 over 6 b with rightwards arrow on top 

Sehingga, DE with rightwards arrow on top equals 1 over 6 a with rightwards arrow on top plus 1 over 6 b with rightwards arrow on top.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Diketahui vektor posisi , , dan , Titik P berada di AB dengan perbandingan , titik Q berada di AC dengan perbandingan , titik R merupakan perpotongan ruas garis BQ dan CP, tentukan: a. koordinat titi...

Pembahasan Soal:

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Roboguru

Diketahui titik-titik . Tentukan: vektor

Pembahasan Soal:

Diketahui size 14px P begin mathsize 14px style left parenthesis 3 comma 5 right parenthesis end style size 14px comma size 14px space size 14px Q begin mathsize 14px style left parenthesis 7 comma negative 1 right parenthesis end style size 14px comma size 14px space size 14px R begin mathsize 14px style left parenthesis negative 1 comma 3 right parenthesis end style size 14px comma size 14px space size 14px dan size 14px space size 14px S begin mathsize 14px style left parenthesis 6 comma negative 2 right parenthesis end style

undefined  

Jadi vektor begin mathsize 14px style stack R S with rightwards harpoon with barb upwards on top equals open parentheses table row 7 row cell negative 5 end cell end table close parentheses end style 

0

Roboguru

Pembahasan Soal:

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Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

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