Reducible to homogeneous differential equation

Question

While studying the cases that are reducible to homogeneous differential equation I have the following issue.

Given $ y = f(\frac{ax+by+c}{a'x+b'y+c})$. It is also given that they cross each other in the point $(x^{}_{0},y^{}_{0})$.

Apply the following transformation

\begin{cases} x = t+x^{}_{0} \\ y = u + y^{}_{0} \end{cases}

According to the book we would get that $ax+by+c$ is equal to $at+bu$ but I don't really get how.

Can someone help please?

Thanks in advance.

Answer 1

If they intersect in $(x_0,y_0)$ then clearly $(x_0,y_0)$ satisfies $ax+by+c=0$ which means: $$ax_0+by_0+c=0 \iff c = -ax_0-by_0$$ Now plug in the given transformation and use that $c = -ax_0-by_0$.

$\begin{cases} x = t+x^{}_{0} \\ y = u + y^{}_{0} \end{cases}$