A line perpendicular to the line segment joining the points (1, 0) and (2, 3)

According to the section formula, the coordinates of the point that divides the line segment joining the points (1, 0) and (2, 3) in the ratio 1: n is given by

$\left(\frac{n(1)+1(2)}{1+n}, \frac{n(0)+1(3)}{1+n}\right)=\left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)$

The slope of the line joining the points (1, 0) and (2, 3) is

$m=\frac{3-0}{2-1}=3$

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points $(1,0)$ and $(2,3)=-\frac{1}{m}=-\frac{1}{3}$

Now, the equation of the line passing through $\left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)$ and whose slope is $-\frac{1}{3}$ is given by

$\left(y-\frac{3}{n+1}\right)=\frac{-1}{3}\left(x-\frac{n+2}{n+1}\right)$

$\Rightarrow 3[(n+1) y-3]=-[x(n+1)-(n+2)]$

$\Rightarrow 3(n+1) y-9=-(n+1) x+n+2$

$\Rightarrow(1+n) x+3(1+n) y=n+11$