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Tentukan koordinat titik P apabila membagi AB dengan perbandingan berikut: c. A ( − 2 , 3 ) , B ( 4 , 1 ) , dan AP : PB = 3 : − 1

Tentukan koordinat titik  apabila begin mathsize 14px style straight P end style membagi  dengan perbandingan berikut:

c. , dan  

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Y. Fathoni

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

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Pembahasan

Gunakan konsep perbandingan vektor. Ingat bahwa. . Perhatikan perhitungan berikut.

Gunakan konsep perbandingan vektor.

Ingat bahwa.

begin mathsize 14px style straight A open parentheses x subscript straight A comma space y subscript straight A close parentheses comma space straight B open parentheses x subscript straight B comma space y subscript straight B close parentheses rightwards arrow AB open parentheses table row cell x subscript straight B minus x subscript straight A end cell row cell y subscript straight B minus y subscript straight A end cell end table close parentheses end style 

begin mathsize 14px style Misal space titik space straight P open parentheses x subscript straight P comma space y subscript straight P close parentheses end style.

Perhatikan perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell AP space colon space PB end cell equals cell 3 space colon space minus 1 end cell row cell AP over PB end cell equals cell fraction numerator 3 over denominator negative 1 end fraction end cell row AP equals cell negative 3 cross times PB end cell row cell open parentheses table row cell x subscript straight P minus x subscript straight A end cell row cell y subscript straight P minus y subscript straight A end cell end table close parentheses end cell equals cell negative 3 cross times open parentheses table row cell x subscript B minus x subscript straight P end cell row cell y subscript B minus y subscript straight P end cell end table close parentheses end cell row cell open parentheses table row cell x subscript straight P minus open parentheses negative 2 close parentheses end cell row cell y subscript straight P minus 3 end cell end table close parentheses end cell equals cell negative 3 cross times open parentheses table row cell 4 minus x subscript straight P end cell row cell 1 minus y subscript straight P end cell end table close parentheses end cell row cell open parentheses table row cell x subscript straight P plus 2 end cell row cell y subscript straight P minus 3 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 12 plus 3 x subscript straight P end cell row cell negative 3 plus 3 y subscript straight P end cell end table close parentheses end cell row cell table attributes columnalign right center left columnspacing 2px end attributes row cell x subscript straight P plus 2 end cell equals cell negative 12 plus 3 x subscript straight P end cell row cell x subscript straight P minus 3 x subscript straight P end cell equals cell negative 12 minus 2 end cell row cell negative 2 x subscript straight P end cell equals cell negative 14 end cell row cell x subscript straight P end cell equals 7 end table end cell blank cell table attributes columnalign right center left columnspacing 2px end attributes row cell y subscript straight P minus 3 end cell equals cell negative 3 plus 3 y subscript straight P end cell row cell y subscript straight P minus 3 y subscript straight P end cell equals cell negative 3 plus 3 end cell row cell negative 2 y subscript straight P end cell equals 0 row cell y subscript straight P end cell equals 0 end table end cell end table end style  

begin mathsize 14px style Jadi comma space diperoleh space titik space straight P open parentheses 7 comma space 0 close parentheses. end style

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Devi Wulandari

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