Using Algebra Expressor (Graphing Software) we can contruct different
graphs for the equation
and we can overlay such graphs for different values of a, b, and c as
the other two are held constant. From these graphs a pattern will develop.
If we set y = 0 we get the equation 0 = x + c. If we graph this equation
for the previous values of c, we get a set of vertical lines parallel to
the y axis ( x = 0 ).
If this equation is graphed in the xc plane we get a line through the origin
(x = y). Look at the above graph; what can be conclustion can be made if
you overlay the graph of x = y, and x = -y on the above graph?
What if we set x = 0, we get the equation 0 = y + c. What do the different
graphs of this equation look like? Again, what conclustions can be made?
It might be helpful to look at the graphs of x - b = 0 and
y - a = 0
The following graph is that of x - b = 0
For the graph of y-a = 0 we get a similar picture, but of course we get
horizontal intead of vertical lines, but the line y = a looks just like
the line x = b.
Now if we look at the graph of the equation (x- b)(y - a) = 0.
The two staight lines is the graph of (x+3)(y-1) = 0. Review the graphs
of x + 3 = 0,
and y - 1 = 0 to see how this graph is obtained. It appears that this equation
forms the asymtopes for the equation xy = x - 3y (a = 1 and b = -3 as in
the other equation).
It might be helpful to look at another picture overlaying these 2 equations
with a different value for b to help us make conclusions. The following
picture is obtained, if we set b=2.
Again, it appears that we are getting the asymptopes.
Again, it appears that the equation (x- 1)(y+3) = 0 forms the asymtopes
for the equation xy = -3x + y (a = -3 and b = 1 as in the other equation).
Therefore it looks like if we set both of these equations for the same values
for a and b keeping c constant, we will get the asymptopes.
Look at the graphs in which c was varied at the beginning. This graph
looks very similar.
Lets see if we can figure out what is going on.
If we take the equation (x - b)(y - a) = k, and multiply we get
xy = ax + by - ab + k.
Now if we compare this to our equation xy = ax + by + c we will get the
same equation for c = -ab + k.
They are the same.