tentukan persamaan garis yang melalui titik (2,6) dan (-1,-4)? pake caranya ya

y-6/-4-6=x-2/-1-2
y-6/-10=x-2/-3
-3(y-6)=-10(x-2)
-3y+18=-10x+20
-3y=-10x+20-18
-3y=-10x+2
-------------------kalikan-1
3y=10x-2

y-y1 per y2-y1 = x-x1 per x2-x1
y-6 per -4-6 = x-2 per -1-2
y-6 per -10 =x-2 per -3
-3(y-6) = -10(x-2)
-3 y + 18 = -10x+20
10x -3y -2 =0