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Persamaan lingkaran yang melalui titik ( 3 , 5 ) , ( − 4 , − 2 ) , dan ( 3 , − 1 ) adalah ...

Persamaan lingkaran yang melalui titik  adalah ...

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S. Yoga

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Mahasiswa/Alumni Universitas Pendidikan Indonesia

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persamaan lingkarannya adalah .

persamaan lingkarannya adalah begin mathsize 14px style x squared plus y squared plus 2 x minus 4 y minus 20 equals 0 end style.

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Melalui titik : ...(1) Melalui titik : ...(2) Melalui : ...(3) Dengan metode substitusi, didapat: Didapat: . Jadi, persamaan lingkarannya adalah .

Melalui titik begin mathsize 14px style left parenthesis 3 comma space 5 right parenthesis end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 squared plus 5 squared plus A times 3 plus B times 5 plus C end cell equals 0 row cell 9 plus 25 plus 3 A plus 5 B plus C end cell equals 0 row cell 3 A plus 5 B plus C plus 34 end cell equals 0 row cell 3 A plus 34 end cell equals cell negative open parentheses 5 B plus C close parentheses end cell row cell 3 A end cell equals cell negative 5 B minus C minus 34 end cell row A equals cell fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction end cell end table end style ...(1)

Melalui titik begin mathsize 14px style left parenthesis negative 4 comma space minus 2 right parenthesis end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared plus A left parenthesis negative 4 right parenthesis plus B left parenthesis negative 2 right parenthesis plus C end cell equals 0 row cell 16 plus 4 minus 4 A minus 2 B plus C end cell equals 0 row cell negative 4 A minus 2 B plus C plus 20 end cell equals 0 end table end style...(2)

Melalui begin mathsize 14px style left parenthesis 3 comma space minus 1 right parenthesis end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 3 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus A left parenthesis 3 right parenthesis plus B left parenthesis negative 1 right parenthesis plus C end cell equals 0 row cell 9 plus 1 plus 3 A minus B plus C end cell equals 0 row cell 3 A minus B plus C plus 10 end cell equals 0 end table end style...(3)

Dengan metode substitusi, didapat: 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Substitusi thin space A end cell equals cell fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction end cell row cell negative 4 times fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction minus 2 B plus C plus 20 end cell equals cell 0... left parenthesis 2 right parenthesis end cell row cell negative fraction numerator 4 open parentheses negative 5 B minus C minus 34 close parentheses over denominator 3 end fraction minus 2 B plus C plus 20 end cell equals 0 row cell fraction numerator 14 B plus 7 C plus 136 over denominator 3 end fraction plus 20 end cell equals cell 0... left parenthesis 4 right parenthesis end cell row blank blank blank row cell 3 times fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction minus B plus C plus 10 end cell equals cell 0 space... left parenthesis 3 right parenthesis end cell row cell open parentheses negative 5 B minus C minus 34 close parentheses minus B plus C plus 10 end cell equals 0 row cell negative 5 B minus C minus 34 minus B plus C plus 10 end cell equals 0 row cell negative 6 B minus 24 end cell equals 0 row B equals cell negative 4 end cell row blank blank blank row cell Substitusi thin space B end cell equals cell negative 4 end cell row cell fraction numerator 14 open parentheses negative 4 close parentheses plus 7 C plus 136 over denominator 3 end fraction plus 20 end cell equals cell 0 space... left parenthesis 4 right parenthesis end cell row cell fraction numerator negative 14 times thin space 4 plus 7 C plus 136 over denominator 3 end fraction plus 20 end cell equals 0 row cell fraction numerator 7 C plus 80 over denominator 3 end fraction plus 20 end cell equals 0 row cell fraction numerator 7 C plus 80 over denominator 3 end fraction end cell equals cell negative 20 end cell row cell 7 C plus 80 end cell equals cell negative 60 end cell row cell 7 C end cell equals cell negative 140 end cell row C equals cell negative 20 end cell row blank blank blank row cell Substitusi thin space C end cell equals cell negative 20 comma thin space B equals negative 4 end cell row A equals cell fraction numerator negative 5 open parentheses negative 4 close parentheses minus open parentheses negative 20 close parentheses minus 34 over denominator 3 end fraction... left parenthesis 1 right parenthesis end cell row A equals cell fraction numerator 5 times thin space 4 plus 20 minus 34 over denominator 3 end fraction end cell row A equals cell 6 over 3 end cell row A equals 2 end table end style 

Didapat: begin mathsize 14px style A equals 2 comma thin space C equals negative 20 comma thin space B equals negative 4 end style.

Jadi, persamaan lingkarannya adalah begin mathsize 14px style x squared plus y squared plus 2 x minus 4 y minus 20 equals 0 end style.

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