Diketahui garis 4x-ay=5 dan 3x +(a+1)y=10 saling tegak lurus. Tentukan;

Garis 4x-ay=5
==> ay = 4x – 5
==> y = 4x/a – 5/a
gradien : 4x/a

garis 3x +(a+1)y=10
==> (a+1)y = -3x + 10
==> y = -3x/(a+1) + 10/(a+1)

diketahui dua garis di atas saling tegak lurus, mka
4/a × -3/(a+1) = -1
-12/a(a+1) = -1
12 = a(a+1)
a² + a = 12
a² + a – 12 = 0
(a + 4)(a – 3) = 0
a = -4   atau   a = 3
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a. titik potong kedua garis adalah
4x/a – 5/a = -3x/(a+1) + 10/(a+1)
misal a = 3
4x/3 – 5/3 = -3x/4 + 10/4
4x/3 + 3x/4 = 5/3 + 10/4
25x/12 = 25/6
x = (25/6)(12/25)
x = 12/6
x = 2

saat x = 2 maka 
y = 4(2)/3 – 5/3
y = 8/3 – 5/3
y = 3/3
y = 1

b. pers. garis yang melalui (0,0) dan (2,1)
m = 1/2
y – 1 = (1/2)(x – 2)
y = x/2 – 1 + 1
y = x/2
2y = x 
2y – x = 0