Persamaan garis yang melalui titik (3,-2) dg gradien 1/2 adl

y-y1 = m(x-x1)
y+2 = 1/2(x-3)
2y+4 = x-3
x - 2y - 7 = 0

langsung aja y

y - y1 = m(x-x1)

y - (-2) = ½ (x - 3)

y + 2 = ½ x - 3/2 (×2)

2y + 4 = x - 3

x - 2y - 3 - 4 = 0

x - 2y - 7 = 0

semoga berguna +_+