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Tentukan koordinat titik P apabila membagi AB dengan perbandingan berikut: a. A ( 1 , 2 ) , B ( 11 , 7 ) , dan AP : PB = 2 : 3

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

a. , dan 

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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 2 space colon space 3 end cell row cell AP over PB end cell equals cell 2 over 3 end cell row AP equals cell 2 over 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 2 over 3 cross times open parentheses table row cell x subscript straight B minus x subscript straight P end cell row cell y subscript straight 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 1 end cell row cell y subscript straight P minus 2 end cell end table close parentheses end cell equals cell 2 over 3 cross times open parentheses table row cell 11 minus x subscript straight P end cell row cell 7 minus y subscript straight P end cell end table close parentheses end cell row cell 3 cross times open parentheses table row cell x subscript straight P minus 1 end cell row cell y subscript straight P minus 2 end cell end table close parentheses end cell equals cell 2 cross times open parentheses table row cell 11 minus x subscript straight P end cell row cell 7 minus y subscript straight P end cell end table close parentheses end cell row cell open parentheses table row cell 3 x subscript straight P minus 3 end cell row cell 3 y subscript straight P minus 6 end cell end table close parentheses end cell equals cell open parentheses table row cell 22 minus 2 x subscript straight P end cell row cell 14 minus 2 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 3 x subscript straight P minus 3 end cell equals cell 22 minus 2 x subscript straight P end cell row cell 3 x subscript straight P plus 2 x subscript straight P end cell equals cell 22 plus 3 end cell row cell 5 x subscript straight P end cell equals 25 row cell x subscript straight P end cell equals 5 end table end cell blank cell table attributes columnalign right center left columnspacing 2px end attributes row cell 3 y subscript straight P minus 6 end cell equals cell 14 minus 2 y subscript straight P end cell row cell 3 y subscript straight P plus 2 y subscript straight P end cell equals cell 14 plus 6 end cell row cell 5 y subscript straight P end cell equals 20 row cell y subscript straight P end cell equals 4 end table end cell end table end style  

begin mathsize 14px style Jadi comma space diperoleh space titik space straight P open parentheses 5 comma space 4 close parentheses. end style

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Aziza ramadhania

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Perhatikan gambar berikut. AP , PQ ​ dan CR masing-masing adalah garis berat segitiga ABC . Buktikan bahwa: b. ∣ ∣ ​ BO ∣ ∣ ​ ÷ ∣ ∣ ​ OQ ​ ∣ ∣ ​ = 2 ÷ 1

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