If a square is formed by x + 3y +4 = 0, 2x + 6y + 7 = 0, 3x – y + 4 = 0 and 6x – y + a = 0, what is the value of a?

To determine a if  the square is formed by x + 3y +4 = 0, 2x +
6y + 7 = 0, 3x – y + 4 = 0 and 6x – y + a = 0, what is the value of
a:

The lines x+3y+4 = 0 and  3x-y+4 are perpendicular as the their
slope is -1/3 and 3 and the product  slope is -1.

Therefore the
other pair of lines 2x+6y +7 = 0 and 6x-y +a = 0 should be perpendicular. The slope of them are
-2/6 and 6/1. The product of the slopes = -(2/6)(6/1) = -2. For the lines to be perpendicular the
product of the slopes must be equal to -1.

For a square to be
formed  any two adjacent lines  equations   must be perpendicular. Therefore the given four lines
does not form a square.

Correcting the equation 6x-y + a = 0 as
6x-2y +a = 0, we get the pars of opposite lines:

x+3y+4 = 0 ||
2x+6y+7 = 0 , Or x+3y+3.5 = 0

3x-y +4 = 0 ||  6x-2y +a = 0. Or 3x-y
+a/2.

The distance between first pair of parallel lines  = {-4
-(-7)}/(3^2+1^2) = 3/sqrt10....................(1)

The distant
between the other pair of || lines = |4- a/2|/sqrt(3^2+1^2) =
(4-a/2)/sqrt10...........(2).

For a square, the distance beween the
opposite pairs of lines must be same.

Therefore 3/sqrt10  =
(4-a/2)/sqrt10.

Or 3 = (4-a/2).

Or a/2
= 4-3 = 1

 a=
2.

Therefore  the value of a = 2 int he equation
6x-2y +a  = 0 in order that the system of equations,

 x + 3y +4 = 0,
2x + 6y + 7 = 0, 3x – y + 4 = 0 and 6x – 2y +
a = 0 should form a square.