persamaan garis G yang melalui titik A(-6,1)dan tegak lurus degan garis yang melalui tirik P(2,-7)dan Q(-3,8)

Kelas 8 Matematika
Bab Persamaan Garis

m1 = (y2 - y1)/(x2 - x1)
m1 = (8 - (-7)) / (-3 - 2)
m1 = 15/(-5)
m1 = -3

Tegak lurus

m1 . m2 = -1
(-3) . m2 = -1
m2 = -1/-3
m2 = 1/3

y - y1 = m (x - x1)
y - 1 = (1/3) . (x + 6)
3 (y - 1) = x + 6
3y - 3 = x + 6
x - 3y + 6 + 3 = 0
x - 3y + 9 = 0

≡ penyelesaian :

⇒Tentukan gradien :

( 2 , - 7 ) = ( x₁ , y₁ )
( - 3 , 8 ) = ( x₂ , y₂ )

↬m₁ = y₂ - y₁ / x₂ - x₁
↬m₁ = 8 - ( -7 ) / - 3 - 2
↬m₁ = 8 + 7 / - 5
↬m₁ = 15 / - 5
↬m₁ = - 3

⇒Karena tegak lurus, maka :

m₁ × m₂ = - 1
- 3 × m₂ = - 1
m₂ = - 1 / - 3
m₂ = ⅓

⇒persamaan garis melalui ( - 6 , 1 ) :

( - 6 , 1 ) = ( x₁ , y₁ )

y - y₁ = m₂ ( x - x₁ )
y - 1 = ⅓ ( x - ( - 6 ) )
y - 1 = ⅓ ( x + 6 )
y - 1 = ⅓x + 2
-------------------- ( ×3 )

3( y - 1 ) = 3( ⅓x + 2 )
3y - 3 = x + 6
x - 3y + 3 + 6 = 0
x - 3y + 9 = 0 [tex]~[/tex]