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Verified answer
A. (y-y1)=m(x-x1)y+8=4(x-3)
y+8=4x-12
y=4x-12-8
y-4x=-20
b. y+4=-⅔(x+6)
(y+4)*3=-2(x+6)
3y+12=-2x-12
3y+2x=-12-12
3y+2x=-24
melalui titik (3, -8)
m = 4
y - y1 = m(x - x1)
y - (-8) = 4(x - 3)
y + 8 = 4x - 12
y = 4x - 12 - 8
y = 4x - 20
atau
4x - y - 20 = 0
3.b)
melalui titik (-6, -4)
m = -2/3
y - y1 = m(x -x1)
y - (-4) = -2/3(x - (-6))
y + 4 = -2/3(x + 6)
-------------------------- kedua ruas dikali 3
3(y + 4) = -2(x + 6)
3y + 12 = -2x - 12
3y = -2x - 12 - 12
3y = -2x - 24
2x + 3y + 24 = 0