Question 17135
  let ABCD be the square. let AB be given by 5x+12y-10=0 .the other equation given is  5x+12y+29=0 ...since ratios of x and y terms in both the lines are same (5/12 and 5/12),they represent parallel sides .so 5x+12y+29=0 represents the equation of CD.
 since AD and BC should be perpendicular to the above lines ,their equations will be of the form 12x-5y+p=0 and 12x-5y+q=0.we are given that 3,5 lies on one of these .hence
  12*3-5*5+p=0 or 36-25+p=0 or p=-11.
hence equation of AD or BC is 12x-5y-11=0 say let it be called AD.
to find the other equation , take any point on CD and find its distance from AB to determine the side of the square.
eqn.of CD is 5x+12y+29=0..take y=0 and then 5x+29=0 or x=29/5 ..so 29/5,0 is a point on CD.its distance from AB is given by 
 Side of square = [5*29/5+12*0-10]/{{{sqrt(5^2+12^2)}}}=19/13
  now BC should be at the same distance from AD.we know already that 3,5 is a point on AD as given in the problem.So let us find distance of 3,5 from BCand equate it to side of square = 19/13
  19/13=(12*3-5*5+q)/{{{sqrt(5^2+12^2)}}}=(11+q)/13(WE MAY TAKE BOTH PLUS AND MINUS SIGNS ON ONE SIDE OF THIS EQUATION TO GET THE 2 EQNS.OF BC WHICH COULD BE ON EITHER SIDE OF AD.)
11+q=19 or 11+q=-19
q=8 or -30 
hence eqns.of BC are 12x-5y+8=0 or 12x-5y-30=0