Suppose that the points

$\because(h, k),(1,2)$ and $(-3,4)$ are collinear

$\therefore\left|\begin{array}{lll}h & k & 1 \\ 1 & 2 & 1 \\ -3 & 4 & 1\end{array}\right|=0 \Rightarrow-2 h-4 k+10=0$

$\Rightarrow h+2 k=5$ .....(i)

Now, $m_{L_{1}}=\frac{4-2}{-3-1}=-\frac{1}{2} \Rightarrow m_{L_{2}}=2 \quad\left[\because L_{1} \perp L_{2}\right]$

By the given points $(h, k)$ and $(4,3)$,

$m_{L_{2}}=\frac{k-3}{h-4} \Rightarrow \frac{k-3}{h-4}=2 \Rightarrow k-3=2 h-8$

$2 h-k=5$ ...(ii)

From (i) and (ii)

$h=3, k=1 \Rightarrow \frac{k}{h}=\frac{1}{3}$