Answers

  • Brainly User
2016-01-29T12:48:57+08:00
1.)  As Parallel:  The slopes must be the same, and the y-intercepts different.
      Transform to slope-intercept form, y = mx + b
       
      Equation 1:
      9x - 6y + 2 = 0
      -6y = -9x - 2
      -6y/-6 = -9x/-6 - 2/-6
       y =  \frac{3x}{2}  \frac{1}{3}
       m = 3/2

      Equation 2:
      3x + ky - 6 = 0
      ky = - 3x + 6
      ky/k = -3x/k + 6/k
      y = - \frac{3x}{k}  \frac{6}{k}
      m = - \frac{3}{k}

     Solve for k to make the system parallel:
     -  \frac{3}{k} =  \frac{3}{2}
       3k = (-3) (2)
       3k = -6
       3k/3 = -6/3
       k = -2
        
Therefore, Equation 2 is :
     3x -2y - 6 = 0
     -2y = -3x + 6
     -2y/-2 = -3x/-2 + 6/-2
      y =  \frac{3x}{2} - 3
      m = 3/2   (same slope as Equation 1)
    
 2.)  As perpendicular, the product of the slopes is -1.  

       Equation 1:
       m = 3/2
           
       Equation 2:
       m = -2/3    
    
       (3/2) (-2/3) = -1  ⇒  Product of slopes is -1.
      
       Solve for k to make the the two functions perpendicular to each other:
       3x + ky - 6 = 0
       ky = -3x + 6
       ky/k = -3x/k + 6/k
       y = - \frac{3}{k} +  \frac{6}{k}
       m = - \frac{3}{k}
     
       Since the slope of the second equation must be - \frac{2}{3} , then:
       
      -3/k = -2/3
      -2k = (-3) (3)
      -2k = -9
      -2k/-2 = -9/-2
       k = 9/2

       To make the lines perpendicular to each other, Equation 2 must be:
        3x + ky - 6 = 0
        3x +  \frac{9y}{2} - 6 = 0
         \frac{9y}{2} = -3x + 6
        
       2/9 (9y/2 = -3x + 6) 2/9
        y = - \frac{2x}{3}  + \frac{4}{3}
        m = -2/3

  ANSWER:  For parallel equations, k = -2.
                      For perpendicular equations, k = 9/2 


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