Jawaban:

1. a) titik (3,4) dan (5, 12)

y = m(x - x1) + y1

y = (12-4)/(5-3) (x - 3) + 4

y = 8/2 (x - 3) + 4

y = 4 (x - 3) + 4

y = 4x -12 + 4

y = 4x - 8

b) titik (6, -8) dan titik (-2,6)

y = m(x - x1) + y1

y = (6+8)/(-2-6) (x - 6) - 8

y = (14/-8) (x -6) - 8

y = - 7/4 (x - 6) - 8

4y = - 7(x- 6) - 32

4y = -7x + 42 -32

4y = -7x + 10

7x + 4y -10 = 0

2. 3x - 2y + 1= 0

2y=3x+1

y = 3/2 x + 1/2

m= 3/2

sejajar: m1=m2

y - y1 = m (x - x1)

y-(-4) = 3/2(x - 1)

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

dikali 2, agar bentuk pecahannya hilang

2y+ 8 = 3x - 3

3x - 2y - 11 = 0

3. cari gradien (m) dari garis x - 2y + 3 = 0

m = -1/-2

m = 1/2

karena tegak lurus maka :

m1.m2 = -1

1/2.m2 = -1

m2 = -1: 1/2

m2 = -2

persamaannya :

y - y1 = m(x - x1) y - 1 = -2(x - 5)

y = -2x + 10 + 1

2x + y - 11 = 0