If two of the adjacent sides of a rectangle have equations x+3y = 8 and ax + y = 4, how can you find a?

The two adjacent sides of a rectangle 
are perpendicular.

The  product of the slopes of two
straight lines  should be -1. These properties are sufficient to determine the
a  in the equations  x+3y = 8 and
ax + y = 4 which represent two adkacent sides of a
rectangle.

Slope of the line, ax^2+bx+c = 0 is
-a/b.

Therefore the slope of the line  x+3y = 8  is -1/3
and  the line ax+ + y = 4 is 
-a/1 =
-a.

Threfore the product of
the slopes = -1. Or

(-1/3)(-a) =
-1.

a/3 = -1.

a
= -3.

Therefore the second equation , ax+y =
4 should be -3x+y =4