Question

Persamaan garis yg melalui titik (6,-1) dan tegak lurus dg garis y=3x+2=....​

Answer 1

y = 3x + 2

y = mx + c

m1 = 3

m x m1 = -1

m= -1/3

y-y1=m(x-x1)

y+1=-1/3(x-6)

y= -1/3x+2-1

y=-1/3x+1

3y=1-x

Answer 2

=>> y = 3x+2 atau 3x-y+2 = 0

gradien atau m1 = 3

didapat:

a = 3 dan b = -1

krn tegak lurus, maka :

m2 = b/a = -1/3

=>> Jadi, persamaan garis melalui titik (6,-1) dan gradien m2 = -1/3 :

y-y1 = m2.(x-x1)

y-(-1) = -1/3 . (x-6)

y+1 = -1/3 . (x-6)

kdua ruas dikalikan dg 3

3.(y+1) = -1.(x-6)

3y+3 = -x+6

x+3y+3-6 = 0

x+3y-3 = 0

atau

x+3y = 3