Question

Persamaan garis melalui titik (-3,11) dan bergradien 3 adalah....

Answer 1

Kalo persamaan garisnya sejajar
y-b = m(x-a)
y-11 = 3(x-(-3))
y-11 = 3x+9
y-3x = 11+9
y-3x = 20 atau
-y+3x = -20

kalo persamaan garisnya tegak lurus
y-b = m(x-a)
y-11 = -1/3(x-(-3))
y-11 = -1/3x-1 [x3]
3y-33 = -x-3
3y+x = 33-3
3y+x = 30 atau
-3y-x = -30

Answer 2

M=gradien

y-y1 = m (x-x1)
y-11= 3(x--3)
y-11= 3(x+3)
y-11= 3x+9
y=3x+9+11
y=3x+20