1.garis melalui titik (9,-5)dengan gradien -3/7 tentukan persamaan garisnya! 2.garis yang melalui titik (4,-21)dan tegak lurus dengan garis y= -3x + 10.tentuka

1) pers garis
y - y1 = m ( x - x1)
y + 5 = - 3/7 ( x - 9)
y = - 3/7 x + 27/7 - 5 (dikali 7)
7y = - 3x + 22 - 35
7y + 3x = - 13

2) y = - 3x + 10
m = - 3
gradien saling tegak lurus :
m1 . m2 = - 1
- 3 . m2 = - 1
m2 = - 1/-3
m2 = ⅓
pers garis
y - y1 = m ( x - x1)
y + 21 = ⅓ ( x - 4)
y = ⅓x - 4/3 - 21 ( dikali 3)
3y = x - 4 - 63
3y - x = - 67

3) 7x + 5y + 1 = 0
5y = - 7x - 1
m = - 7/5
sejajar m1 = m2 = - 7/5
pers garis
y - y1 = m ( x - x1)
y - 5 = - 7/5 ( x + 6)
y = - 7/5 x - 42/5 + 5 ( dikali 5)
5y = - 7x - 42 + 25
5y + 7x = - 17

4) 3y + x - 5 = 0
3y = - x + 5
m = - ⅓
sejajar m1 = m2 = - ⅓
pers garis
y - y1 = m ( x - x1)
y - 6 = - ⅓ ( x + 4)
y = - ⅓ x - 4/3 + 6 ( dikali 3)
3y = - x - 4 + 6
3y + x = 2