1. persamaan garis lurus yang sejajar dengan garis 2x+3y+6=0 melalui titik (-2,5) . jawab 2. persamaan garis lurus yang melalui titik A (-2,3) dan tegak lurus t

1.2x+3y+6=0
 3y = -2x-6
 y = -2/3x -6/3
 y= -2/3x -2

m = -2/3 c=-2

y-y1= m(x-x1)
y-5 = -2/3 (x-(-2)) -> Kali silang
3(y-5)= -2(x+2)
3y-15 = -2x-2
3y      = -2x-2+15
3y      = -2x+13 
y        = -2/3 x + 13/3


2. m1 = 3/2

  kalau sifat nya tegak lurus m1xm2= -1
      3/2 x m2 =-1
              m2 = -1 x 2/3
              m2 = -2/3 

y-y1 = m(x-x1)
y-3  = -2/3 (x-(-2)) -> Kali silang
3(y-3) = -2(x+2)
3y-9  =-2x-4 
3y     = -2x-4+9 
3y     = -2x + 5
y        = -2/3+ 5/3

3.[tex] \frac{y-y1}{y2-y1}= \frac{x-x1}{x2-x1} \\ \frac{y-7}{-1-7}= \frac{x-(-4)}{10-(-4)} \\ \frac{y-7}{-8} = \frac{x+4}{14} (Kalisilang) \\ 14y-98= -8x-32 \\ 14y = -8x-32+98 \\ 14y= -8x + 66 \\ y= -8/14x + 66/14[/tex]

Kalau salah maaf semoga membantu :)