Slope of the line joining the points $A(1,0)$ and $B(2,3)$
$=\frac{3-0}{2-1}=\frac{3}{1}=3$
$\therefore \quad$ Slope of the $C D$ perpendicular to
$A B=-\frac{1}{3}$


$[\because$ product of slopes of perpendicular lines $=-1]$
The point $P$ divides $A B$ in the ratio $1: n$ $\therefore \quad$ Coordinates of $P$ are,
$\left(\frac{1 \times 2+1 \times n}{1+n}, \frac{1 \times 3+0 \times n}{1+n}\right) \text { or }\left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)$
Equation of the line $C D$ which is perpendicular to $A B$ passes through $P$ is
$y-\frac{3}{n+1}=-\frac{1}{3}\left(x-\frac{n+2}{n+1}\right)$
$\left[\right.$ Using the formula $\left.y-y_1=m\left(x-x_1\right)\right]$
$\begin{aligned} & 3(n+1) y-9=-(n+1) x+(n+2) \\ \Rightarrow \quad &(n+1) x+3(n+1) y=n+2+9 \\ \Rightarrow \quad &(n+1) x+3(n+1) y=n+11 \end{aligned}$