Given point is (3,2)

Equation of line is :  x-2y = 3

 => x-3 = 2y

=> y = x/3 - 1 

Let the slope of required line is m.

Again given that slope between reqired line and x - 2y = 3

tan45 = | (1/2 - m)/(1+m/2) |

=> 1 =| (1/2 - m)/(1+m/2) |

Now 

      1 = (1/2 - m)/(1+m/2)  and -1= (1/2 - m)/(1+m/2)    (since |x| = -x and x)

=> 1+ m/2 = 1/2 - m and -(1+ m/2) = (1/2 - m)

=> m + m/2 = 1/2 -1 and -1 - m/2 = 1/2 - m

=> 3m/2 = -1/2 and m - m/2 = 1+ 1/2

=> m = -1/3 and m/2 = 3/2

=> m =-1/3 and m = 3

Case1: m = -1/3 and point is (3,2)

      y - 2 = (-1/3)(x-3) 

=> 3(y-2) = -x + 3

=> 3y-6 = -x + 3

=> 3y + x = 3+6

=> x + 3y = 9

Case2: m=3 and point is (3,2)

      y - 2 = 3(x-3)

=> y-2 = 3x - 9

=> 3x - y = 9-2

=> 3x - y = 7