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( x - 2y ) + ax + ay = 5
x + ax - 2y + ay = 5
( 1 + a ) x - ( 2-a) y = 5
maka M = 1 + a
2 - a
garis g :
( 5y - 3x ) - 3ax - 3ay = 12
-3x - 3ax + 5y - 3ay = 12
( -3 - 3a ) x + ( 5 - 3a) y = 12
maka M = 3 + 3a
5 - 3a
a) l // g
maka Ml dan Mg sama
1 + a = 3 + 3a
2 - a 5 - 3a
1 ( 5 - 3a ) + a( 5 - 3a ) = 2( 3 + 3a) - a( 3 + 3a )
5 - 3a + 5a - 3a² = 6 + 6a - 3a - 3a²
5 + 2a = 6 + 3a
a = -1
b ) l tegak lurus g
maka Ml x Mg = -1
1 + a x 3 + 3a = -1
2 - a 5 - 3a
(5y – 3x) –3a(x + y) = 12.
Tentukan nilai a agar:
a.l / / g
b.l⊥ g
JAWAB
◆l: (x – 2y) + a(x + y) = 5
⇨ x - 2y + ax + ay = 5
⇨ (1+a)x + (a-2)y = 5
maka
ml = -(1+a)/(a-2) = (1+a)/(2-a)
◆g: (5y – 3x) –3a(x + y) = 12
⇨ 5y - 3x - 3ax - 3ay = 12
⇨ (5-3a)y - (3+3a)x = 12
maka
mg = -(-3-3a)/(5-3a) = (3+3a)/(5-3a)
a. l / / g, maka
⇨ ml = mg
⇨ (1+a)/(2-a) = (3+3a)/(5-3a)
⇨ (1+a)(5-3a) = (3+3a)(2-a)
⇨ 6a - 5a = 5 - 6
⇨ a = -1
b. l⊥ g, maka
⇨ ml × mg = -1
⇨ (1+a)/(2-a) × (3+3a)/(5-3a) = -1
⇨ (3a²+6a+3)/(3a²-11a+10) = -1
⇨ 3a²+6a+3 = -3a²+11a-10
⇨ 6a² - 5a + 13 = 0
Nilai diskriminan
⇨ D = b² - 4.a.c
⇨ D = (-5)² - 4.6.13
⇨ D = 25 - 312
⇨ D = -287
Nilai D<0 maka persamaan tidak memiliki akar real atau nilai a tidak real, yang artinya kedua garis tidak mungkin tegak lurus.