SOLUTION: two sides of a square have the equation 5x+12y-10=0 and 5x+12y+29=0 the co-ordinates of a point on the third side(3,5).find the equation for the other two sides of square

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Question 17135: two sides of a square have the equation 5x+12y-10=0 and 5x+12y+29=0
the co-ordinates of a point on the third side(3,5).find the equation for the other two sides of square
Answer by venugopalramana(3286) (Show Source):
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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]/=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)/=(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