Tentukan persamaan garis yang sejajar dan tegak lurus melalui titik (-4,-3) dengan persamaan 2y-3x=6

2y-3x=6
kita cari m/Gradiennya
m=-x/y
m= -(-3)/2
m= 3/2

rumus mencari pers garis        --> Sejajar (-4,-3)
y-y1=m(x-x1)
y-(-3)=3/2(x-(-4)
y+3=3/2x+6
y=3/2x+6-3
y=3/2x+3

2y=3x+6
-3x+2y-6=0 (sejajar)

sejajar
2y-3x=6 => 2y=3x+6/2 = y=3/2x+3
(-4,-3)
y-y1=m(x-x1)
y-(-3)=3/2(x-(-4)
y+3=3/2x+6
y=3/2x+6-3 =>y=3/2x+3

tegak lurus
2y-3x=6 => 2y=3x+6/2 = y=3/2x+3 =>-1/3/2=-2/3
(-4,-3)
y-y1=m(x-x1)
y-(-3)-2/3(x-(-4)
y+3=-2/3x-8/3
y=-2/3x-8/3-3 =>y=-2/3x-17/3