# 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 asyntots of our 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?

• I'm not sure to have understood precisely your point. From here we have $$y=\frac{ax+b}{cx+d}=\frac{c(ax+b)}{c(cx+d)}=\frac{a(cx+\frac{bc}a)}{c(cx+d)}$$ which is constant when $\frac{bc}a=d$. – user Jul 31 at 23:01
• @user Not really, I'd like to get the $ad-bc$ condition starting from the $(1)$. – Sebastiano Jul 31 at 23:06

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 $$y=\frac bd+\left(\frac{ax+b}{cx+d}-\frac bd\right)=\frac bd+\frac{(ad-bc)x}{d(cx+d)}.$$