G7.3 The equation of the line through $(4,3)$ and $(12,-3)$ is $\frac{x}{a}+\frac{y}{b}=1$. Find $a$.
$\begin{array}{l}\frac{y-3}{x-4}=\frac{3-(-3)}{4-12} \\ 3 x-12+4 y-12=0 \\ \Rightarrow 3 x+4 y=24 \\ \frac{x}{8}+\frac{y}{6}=1 \\ \Rightarrow a=8\end{array}$