Persamaan garis lurus.

Bila persamaan garis dalam bentuk;

Ax + By + C = 0 ---------> m = -A/B

Syarat 2 garis sejajar --------> m₁ = m₂

syarat 2 garis saling tegak lurus -------> m₁ x m₂ = -1

cara menyusun persamaan garis melalui titik ( x , y ) dan bergradien m;

y - y₁ = m ( x - x₁ )

1.

garis 1.

5x + 2y - 10 = 0

m₁ = -5/2

garis 1 sejajar garis 2, maka m₂ = -5/2

garis 2;

melalui titik( 3 , 5 ) dan m = -5/2

y - 5 = -5/2 ( x - 3 )

2y - 10 = -5 ( x - 3 ) -------> dikalikan 2

2y - 10 = -5x + 15

5x + 2y - 10 - 15 = 0

5x + 2y - 25 = 0

2.

garis 1;

4x - 3y = 12

m₁ = -4/-3 = 4/3

garis 1 sejajr garis 2 maka m₂ = 4/3

garis2.

melalui titik ( -2 , 3 ) dan m = 4/3

y - 3 = 4/3 ( x + 2 )

3y - 9 = 4 ( x + 2 ) ----------> dikalikan 3

3y - 9 = 4x + 8

-4x + 3y = 8 + 9

-4x + 3y = 17

4x - 3y = -17

3.

garis 1.

3y = 7x - 21

-7x + 3y = -21

m₁ = -(-7)/3 = 7/3

garis 1 sejajar garis 2 , maka m₂ = 7/3

garis 2 melalui titik ( -3 , -2 ) dan m = 7/3

y +2 = 7/3 ( x + 3 )

3y + 6 = 7 ( x + 3 )

3y + 6 = 7x + 21

3y = 7x + 21 - 6

3y = 7x + 15

4.

garis 1

5x = 15 - 3y

5x + 3y = 15

m₁ = -5/3

garis 1 sejajar garis 2 , maka m₂ = -5/3

garis 2.

melalui titik ( -7, 0 ) dan m = -5/3

y - 0 = -5/3 ( x + 7 )

3y = -5 ( x + 7 )

3y = -5x - 35

5x = -35 - 3y