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Jika koordinat titik , dan , nilai  adalah ....

Pertanyaan

Jika koordinat titik begin mathsize 14px style text A end text open parentheses m comma negative 5 comma 2 n close parentheses text ,B end text open parentheses n comma 2 m comma m plus 2 close parentheses end style, dan begin mathsize 14px style stack text AB end text with bar on top equals open parentheses table row cell negative 7 end cell row cell 1 minus 4 n end cell row cell 3 m end cell end table close parentheses end style, nilai begin mathsize 14px style m end style adalah ....

  1. begin mathsize 14px style negative 3 end style  

  2. begin mathsize 14px style negative 4 end style  

Pembahasan Soal:

Jika koordinat titik begin mathsize 14px style text A end text open parentheses m comma negative 5 comma 2 n close parentheses text ,B end text open parentheses n comma 2 m comma m plus 2 close parentheses end style, dan begin mathsize 14px style stack text AB end text with bar on top equals open parentheses table row cell negative 7 end cell row cell 1 minus 4 n end cell row cell 3 m end cell end table close parentheses end style, maka

begin mathsize 14px style stack text AB end text with bar on top equals open parentheses table row cell negative 7 end cell row cell 1 minus 4 n end cell row cell 3 m end cell end table close parentheses end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell B minus A end cell equals cell open parentheses table row cell negative 7 end cell row cell 1 minus 4 n end cell row cell 3 m end cell end table close parentheses end cell row cell open parentheses table row n row cell 2 m end cell row cell m plus 2 end cell end table close parentheses minus open parentheses table row m row cell negative 5 end cell row cell 2 n end cell end table close parentheses end cell equals cell open parentheses table row cell negative 7 end cell row cell 1 minus 4 n end cell row cell 3 m end cell end table close parentheses end cell row cell open parentheses table row cell n minus m end cell row cell 2 m plus 5 end cell row cell m minus 2 n plus 2 end cell end table close parentheses end cell equals cell open parentheses table row cell negative 7 end cell row cell 1 minus 4 n end cell row cell 3 m end cell end table close parentheses end cell end table end style 

Dengan menyamakan ruas kiri dan ruas kanan, diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n minus m end cell equals cell negative 7 end cell row cell 2 m plus 4 n end cell equals cell negative 4 end cell row cell negative 2 m minus 2 n end cell equals cell negative 2 end cell end table end style 

Dengan eliminasi diperoleh

begin mathsize 14px style table row cell 2 m plus 4 n equals negative 4 end cell row cell negative 2 m minus 2 n equals negative 2 end cell row cell 2 n equals negative 6 end cell row cell n equals negative 3 end cell end table plus end style 

Substitusikan begin mathsize 14px style n equals negative 3 end style pada persamaan begin mathsize 14px style n minus m equals negative 7 end style. Diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n minus m end cell equals cell negative 7 end cell row cell negative 3 minus m end cell equals cell negative 7 end cell row m equals cell negative 3 plus 7 end cell row m equals 4 end table end style  

Jadi, nilai undefined adalah 4

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 14 Maret 2021

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