Suatu persamaan garis yang melalui titik (4,-3) dan tegak lurus terhadap garis 4y-6x+10=0

garis 4y-6x+10=0
m1=-a/b
m1=-(-6)/4
m1=6/4
m1=3/2

Tegak lurus → m1m2=-1
3/2m2=-1
m2=-2/3
m=-2/3

melalui titik (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-1/3
3y=-2x-1
2x+3y+1=0

Silakan dicek lagi.

4y - 6x + 10 = 0
=> 4y = 6x - 10
=> y = 6/4x - 10/4
m = 6/4

karena tegak lurus maka m = - 4/6
=> y - y1 = m (x - x1)
=> y - (-3) = -4/6 (x - 4)
=> y + 3 = -4/6x + 16/6
=> y = -4/6x + 16/6 - 3
=> y = -4/6x - 2/6