Homographic function: alternative proofs to obtain $ad-bc$ Considering the function,
$$y=\frac{ax+b}{cx+d}\tag1$$
If $c = 0 \wedge d\neq 0$, the function represents a straight line of equation
$$y=\frac ad x+ \frac bd$$
If $c ≠ 0$ and $ad = bc$ the function represents a horizontal straight line. In fact, if
$$ad = bc \tag 2$$
we will have
$$ad/c = bc/c \iff ad/c = b$$
The coordinates of the point $P_0(-d/c,a/c)$ represent the asymptotes of  hyperbola $(1)$. The importance of $(2)$ is due to the reason that if $ad-bc \neq 0$, using the traslation $\tau$,
$$\tau: \begin{cases}
X=x+\dfrac dc & \\
Y=y-\dfrac ac
\end{cases}
$$
I will obtain an equilater hyperbola. In fact
$$Y+\frac{a}{c}=\frac{a\Big(X-\frac{d}{c}\Big)+b}{c\Big(X-\frac{d}{c}\Big)+d}$$
$$Y=\frac{aX-\frac{ad}{c}+b}{cX-d+d}-\frac{a}{c}\Rightarrow Y=\frac{aX-\frac{ad}{c}+b}{cX}-\frac{a}{c}\Rightarrow Y=\frac{aX-\frac{ad}{c}+b-aX}{cX}$$
Hence:
$$Y=\frac{-\frac{ad}{c}+b}{cX}\Rightarrow XY=-\frac{ad}{c^2}+\frac{b}{c}\Rightarrow XY=k$$
with $$k=\frac{bc-ad}{c^2}$$
$$XY=k \tag 3$$

Starting from $(1)$ how can I create the condition quickly (step by step) $$\boxed{\color{orange}{ad-bc}} \quad ?$$
different from my proof?


If $c\neq 0$, then $$y=\frac ac+\left(\frac{ax+b}{cx +d}-\frac ac\right)=\frac ac+\frac{bc-ad}{c(cx +d)}.$$
If $d\neq 0$, then $$y=\frac bd+\left(\frac{ax+b}{cx+d}-\frac bd\right)=\frac bd+\frac{(ad-bc)x}{d(cx+d)}.$$
