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Jika titik A ( 2 x , 3 , 5 ) , B ( 4 , y , 1 ) dan C ( 22 , 5 , 4 ) terletak pada suatu garis lurus (kolinear), tentukan nilai 3 x + 2 y .

Jika titik  dan  terletak pada suatu garis lurus (kolinear), tentukan nilai .

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W. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

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nilai adalah .

nilai begin mathsize 14px style 3 x plus 2 y end style adalah 64.

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Pembahasan

Jika titik , dan terletak pada suatu garis lurus (kolinear), maka: Sehingga: - Menentukan nilai - Menentukan nilai - Menentukan nilai Kemudian: Jadi, nilai adalah .

Jika titik begin mathsize 14px style A left parenthesis 2 x comma 3 comma 5 right parenthesis end stylebegin mathsize 14px style B open parentheses 4 comma y comma 1 close parentheses end style dan begin mathsize 14px style C open parentheses 22 comma 5 comma 4 close parentheses end style terletak pada suatu garis lurus (kolinear), maka:

space space space space space space space space space space space space space space space space space AB equals straight k times BC space space space space space space space space space space space space space straight b with rightwards arrow on top minus straight a with rightwards arrow on top equals straight k times open parentheses straight c with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses open parentheses table row 4 row straight y row 1 end table close parentheses minus open parentheses table row cell 2 straight x end cell row 3 row 5 end table close parentheses equals straight k times open square brackets open parentheses table row 22 row 5 row 4 end table close parentheses minus open parentheses table row 4 row straight y row 1 end table close parentheses close square brackets space space space space space space open parentheses table row cell 4 minus 2 straight x end cell row cell straight y minus 3 end cell row cell negative 4 end cell end table close parentheses equals straight k times open parentheses table row 18 row cell 5 minus straight y end cell row 3 end table close parentheses 

Sehingga:

- Menentukan nilai straight k 

negative 4 equals straight k times 3 straight k times 3 equals negative 4 space space space space straight k equals fraction numerator negative 4 over denominator 3 end fraction 

- Menentukan nilai straight x 

4 minus 2 straight x equals straight k times 18 4 minus 2 straight x equals fraction numerator negative 4 over denominator 3 end fraction times 18 4 minus 2 straight x equals negative 24 space space space minus 2 straight x equals negative 24 minus 4 space space space minus 2 straight x equals negative 28 space space space space space space space space space straight x equals fraction numerator negative 28 over denominator negative 2 end fraction space space space space space space space space space straight x equals 14 

- Menentukan nilai straight y 

space space space space space space space straight y minus 3 equals straight k times open parentheses 5 minus straight y close parentheses space space space space space space space straight y minus 3 equals fraction numerator negative 4 over denominator 3 end fraction open parentheses 5 minus straight y close parentheses space space space space space space space straight y minus 3 equals negative 20 over 3 plus fraction numerator 4 straight y over denominator 3 end fraction space space space space straight y minus fraction numerator 4 straight y over denominator 3 end fraction equals fraction numerator negative 20 over denominator 3 end fraction plus 3 fraction numerator 3 straight y over denominator 3 end fraction minus fraction numerator 4 straight y over denominator 3 end fraction equals fraction numerator negative 20 over denominator 3 end fraction plus 9 over 3 space space space space space space space space fraction numerator negative straight y over denominator 3 end fraction equals fraction numerator negative 11 over denominator 3 end fraction space space space space space space space space space space space space straight y equals 11 

Kemudian:

3 straight x plus 2 straight y equals 3 open parentheses 14 close parentheses plus 2 open parentheses 11 close parentheses equals 42 plus 22 equals 64 

Jadi, nilai begin mathsize 14px style 3 x plus 2 y end style adalah 64.

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