When a Rational function becomes a line?

Question

Why is it that when $AD = BC$, this equation becomes a horizontal line?

$$y = \frac {Ax+B}{Cx+D} $$

For any other values where $AD$ isn't equal to $BC$ it is a rational function.

Answer 1

$$\frac{Ax+B}{Cx+D}=\frac BD\frac{D(Ax+B)}{B(Cx+D)}=\frac BD\frac{ADx+BD}{\color{green}{BC}x+BD}=\frac BD\frac{ADx+BD}{\color{green}{AD}x+BD}=\frac BD.$$ Or $$\frac{Ax+B}{Cx+D}=\frac{Ax+B}{Cx+\dfrac{BC}A}=A\frac{Ax+B}{ACx+BC}=\frac AC\frac{Ax+B}{Ax+B}=\frac AC.$$

Answer 2

You can show it like that :

$$ \frac{Ax+B}{Cx+D} = \frac{A}{C}\frac{Cx + B\frac{C}{A} }{Cx+D}$$ $$ = \frac{A}{C}\frac{Cx + D + (B\frac{C}{A}-D )}{Cx+D}$$ $$ = \frac{A}{C} + \frac{A}{C} \frac{ B\frac{C}{A}-D }{Cx+D}$$ $$ = \frac{A}{C} + \frac{1}{C} \frac{ BC-AD }{Cx+D}$$ $$ = \frac{A}{C} + 0$$

Answer 3

We have $AD = BC$ and $C \ne 0$ or $D \ne 0$, otherwise the expression is not well defined.

Case $C \ne 0$, $D = 0$: This implies $BC = 0$, so we have $B = 0$ and $$ y = \frac{Ax}{C x} = \frac{A}{C} $$ which is a constant function.

Case $C=0$, $D \ne 0$: This implies $AD = 0$, so we have $A=0$ and $$ y = \frac{B}{D} $$ which is a constant function as well.

Case $C \ne 0$, $D \ne 0$: So we have $A = (BC)/D$ and $$ y = \frac{Ax + B}{C x + D} = \frac{((BC)/D) x + B}{Cx + D} = \frac{BC x + BD}{D(Cx + D)} = \frac{B(C x + D)}{D(Cx + D)} = \frac{B}{D} $$ So we always get some constant function.