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y = 3x -2
m1 = 3
syarat garis tegak lurus yaitu m1 . m2 = -1
m1 . m2 = -1
3 . m2 = -1
m2 = -1/3
persamaan garis melalui titik (4,-2) dan gradien m2 = -1/3
y - y1 = m2(x - x1)
y - (-2) = -1/3(x - 4)
3y + 6 = -1(x - 2)
3y + 6 = -x + 2
x + 3y + 4 = 0
3 m2 = -1
m2 = -1/3
persamaa garis :
y = mx + c
y = -1/3 + c
melalui titik (4, -2), artinya :
-2 = -1/3 (4) + c
-2 = -4/3 + c
c = 4/3 - 2
c = -2/3
jadi persamaan garis yg dimaksud adalah y = -1/3 x - 2/3