Jawaban persamaan garis yang melalui titik (-2,3) Dan tegak lurus dengan garis 3y+x-6=0 adalah Lengkap

Jika kamu sedang mencari jawaban atas pertanyaan: persamaan garis yang melalui titik (-2,3) Dan tegak lurus dengan garis 3y+x-6=0 adalah, maka kamu berada di tempat yang tepat. Disini ada beberapa jawaban mengenai pertanyaan tersebut. Silakan baca lebih lanjut.

Pertanyaan

persamaan garis yang melalui titik (-2,3) Dan tegak lurus dengan garis 3y+x-6=0 adalah

Jawaban #1 untuk Pertanyaan: persamaan garis yang melalui titik (-2,3) Dan tegak lurus dengan garis 3y+x-6=0 adalah

pertama cari gradien(m)
m=1/3 karena tegak lurus jadi -3

terakhir cari persamaan garis
y-y1=m(x-x1)
y-3=-3(x+2)
y-3=-3x-6
3x+y-3+6=0
3x+y+3=0

Sekian tanya-jawab mengenai persamaan garis yang melalui titik (-2,3) Dan tegak lurus dengan garis 3y+x-6=0 adalah, semoga dengan ini bisa membantu menyelesaikan masalah kamu.

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