1. Diketahui garis 4x - ay = 5 dan 3x + (a + 1)y = 10 saling tegak lurus. Tentukkan A. Nilai a B. Titik potong kedua garis. C. Persamaan garis yang melalui titu

4x - ay = 5
m₁ = -4/-a = 4/a

3x + (a + 1)y = 10
m₂ = -3/(a + 1)

a. Tegak lurus , m₁ • m₂ = -1
4/a • (-3)/(a + 1) = -1
12/(a² + a) = 1
12 = a² + a
a² + a - 12 = 0
(a + 4)(a - 3) =0
a = -4 atau a = 3

Kita ambil nilai positif saja , yaitu 3

b. 4x - 3y = 5 |x 3|
3x + 4y = 10 |x 4|

Menjadi :
12x - 9y = 15
12x + 16y = 40
-------------------- (-)
-25y = -25
y = 1

4x - 3(1) = 5
4x = 5 + 3
4x = 8
x = 2

Titik Potongnya (2 , 1)

c. Persamaan garis yg melalui (0 , 0) dan (2 , 1)
m = Δy / Δx
m = (1 - 0) / (2 - 0)
m = ½

y - y₁ = m(x - x₁)
y - 1 = ½(x - 2)
2y - 2 = x - 2
x - 2y = 0

-ay = 5-4x
y = -5/a + (4/a)x .....i)
m1 = 4/a

3x+(a+1)y=10
(a+1)y=10-3x
y= 10/(a+1) + (-3/(a+1))x ....ii)
m2= -3/(a+1)

tgk lurus jika
m1 . m2 = -1
4/a . -3/(a+1) = -1
-12/(a(a+1)) = -1
-12 = -1(a(a+1))
12 = a(a+1)
12 = a²+a
a²+a-12 = 0
(a+4)(a-3)=0
a= -4 atau a=3

misal gunakan a=3
dr i) 4x - 3y = 5
ii) 3x + 4y = 10
eliminasi i) x 4 dan ii) x3
16x - 12y =20
9x + 12y = 30
--------------------- +
25x. = 50
x = 2

i) 4(2) -3y = 5
8 -3y = 5
-3y = 5-8 = -3
y = 1
titik potong (2,1)

pg yg mllaui (0,0) dan (2,1)
y-0/1-0 = x-0/2-0
y/1 = x/2
y = (1/2)x