1. persamaan garis a) gradien = 1/2 , melalui titik (2,3) b) gradien -5 , melalui titik (-2,-3) c) melalui titik A(-3,-5) d) melalui titik A(-4,0) , B(5,2) e) m

Jawab :

1. a) m = 1/2 , melalui titik (2,3)
y - y1 = m(x - x1)
y - 3 = 1/2(x - 2)
y - 3 = 1/2x - 1/2 x 2
y - 3 = 1/2x - 1
y = 1/2x - 1 + 3
y = 1/2x + 2

b) m = -5 , melalui titik (-2,-3)
y - y1 = m(x - x1)
y - (-3) = -5(x - (-2))
y + 3 = -5(x + 2)
y + 3 = -5x - 10
y + 3 = -5x - 10 - 3
y = -5x - 13

c) A(-3,-5)
m = y/x = -5/-3 = 5/3
y - y1 = m(x - x1)
y - (-5) = 5/3(x -(-3))
y + 5 = 5/3(x + 3)
y + 5 = 5/3x + 5/3 x 3
y + 5 = 5/3x - 5
y = 5/3x - 5 - 5
y = 5/3x

d) A(-4,0) , dan B(5,2)
m = y2 - y1/ x2 - x1 = 2 - 0/ 5 - (-4) = 2/9
y - y1 = m(x - x1)
y - 0 = 2/9(x -(-4))
y - 0 = 2/9(x + 4)
y - 0 = 2/9x + 2/9 x 4
y - 0 = 2/9x + 8/9
------------------------------ x 9
9.y - 0 . 9 = 2/9x . 9 - 8/9 . 9 

9y - 0 = 2x + 8
9y - 0 - 2x - 8 = 0
9y - 2x - 8 = 0 => -2x + 9y - 8 = 0 

e) A(5,1) , B(-6,-1)
m = y2 - y1/ x2 - x1 = (-1) - 1/ (-6) - 5 = 2/11
y - y1 = m(x - x1)
y - 1 = 2/11(x - 5)
y - 1 = 2/11x - 2/11 x 5
y - 1 = 2/11x - 10/11
--------------------------------- x 11
11.y - 1 . 11 = 2/11x . 11 - 10/11 . 11

11y - 11 = 2x - 10
11y - 11 - 2x + 10
11y - 2x + 21 = 0
-2x + 11y + 21 = 0

2) Menggunakan metode substitusi

a) x + y = 5 , dan 2x - y = 1
x = 5 - y => 2(5 - y) - y = 1
y = 3           10 - 2y - y = 1
x = 5 - y      10 - 3y = 1
x = 5 - 3       -3y = 1 - 10
x = 2            -3y = -9 = y = 3

Jadi titik potongnya : (x,y) = (2,3)