persamaan garis yang melalui titik (-3,-3) dan tegak lurus garis y=-6x + 2 adalah

m = -a                                 m1 x m2 = -1
      -----                                -6  x m2 = -1
        b
    = -(-6)                                    m2  = 1
       ------                                            -----
         -1                                                6
    = 6
      ---
      -1
    = -6

y - y1 = m (x - x1)
y - (-3) =  1 (x - (-3)
               ---
                6
------------------------------  x 6
6y - (-18) = 1 (x + 3)
6y + 18 =  x + 3
-x + 6y +15 = 0

Persamaan Garis

Melalui (-3,-3)
tegaklurus 6x + y - 2 = 0

PG :
1x - 6y = 1(-3) - 6(-3)
x - 6y = -3 + 18
x - 6y = 15 ✔