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Verified answer
M = -3/4 tegak lurus4/3titik yang dilewati (0,3)
y-y1 = m (x-x1)
y-3 = 4/3 (x-0)
y-3 = 4/3x
--------------------×3
3y-9 = 4x
= 4x-3y+9=0
Dicari dulu gradien garis a
ma = -∆Y/∆X
ma = -3/4
Karena garis a tegak lurus dengan garis b, berlaku
ma x mb = -1
-3/4 x mb = -1
mb = -1 : (-3/4)
mb = -1 x (-4/3)
mb = 4/3
Persamaan garis b melalui titik (0, 3) dan bergradien 4/3
y - y1 = m (x - x1)
y - 3 = 4/3 (x - 0)
y - 3 = 4/3x
3y - 9 = 4x
4x = 3y - 9
4x - 3y + 9 = 0
Jawaban: B
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