Question 87791

 Problem 40 requires that you find the slope of the line that is perpendicular to the one represented by the given equation.
L has y-intercept (0, 2) and is perpendicular to the line with equation: 

2x – 3y = 6.

 L is the line for which we have to find the eq'n and given L is perpendicular 
 to 2x-3y = 6, and has intercept (0,2)
  Let the eq'n to L be Y=mx+c  therefore 2 = m.0 +c  or c = 2
    to find the slope of the line 2x-3y = 6 ; -3y = -2x+6 ; y = 2x/3+6/-3
          y = 2x/3-2 ; therefore slope m =2/3 
 the slope of line L = -3/2 (the lines are perpendicular hence product of thier slopes = -1,, M.m = -1, M = -1/2/3 = -3/2)
   the eq'n to the line L is y = -3x/2+2 (by substituting for M & c from the above)  or  eq'n can be reduced to 2y = -3x+4  OR 3x+2y = 4