Given this identity $\delta(kx)=\frac{1}{|k|}\delta(x)$ is the following correct?
$$\delta(ct-x)=\delta(c(t-x/c))=\frac{1}{|c|}\delta(t-x/c)$$
It seems like the impulses would be located at the same point when referred to the x axis but would have different 'areas' because of the 1/c..
You can see that's not the case when you calculate the "areas". In one hand you have (for simplicity I will assume $c > 0$)
$$ \int_{-\infty}^{+\infty}{\rm d}x~\delta(ct - x) = 1 \tag{1} $$
In the other
$$ \int_{-\infty}^{+\infty}{\rm d}x~\frac{1}{c}\delta\left(t - \frac{x}{c}\right) = \int_{\infty}^{+\infty} {\rm d}\left(\frac{x}{c}\right)\delta\left(t - \frac{x}{c}\right) \stackrel{\color{blue}{u}=x/c}{=} \int_{-\infty}^{+\infty} {\rm d}\color{blue}{u}~ \delta(t - \color{blue}{u}) = 1 \tag{2} $$
So the areas are the same