titik a(3 2 -1) b(1 -2 1) c(7 p-1 -5) segaris untuk nilai p=
Matematika
Alvinaputri19
Pertanyaan
titik a(3 2 -1) b(1 -2 1) c(7 p-1 -5) segaris untuk nilai p=
1 Jawaban
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1. Jawaban Adjie564
Syarat suatu garis kolinear
AC = mAB
AB = B - A = (1 -2 1)-(3 2 -1) = (-2 -4 2)
AC = C - A = (7 p-1 -5)-(3 2 -1) = (4 p-3 -4)
(4 p-3 -4) = m(-2 -4 2)
(4 p-3 -4) = (-2m -4m 2m)
4 = -2m
m = -2
p - 3 = -4m = -4(-2) = 8
p = 8+3 = 11