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Q.
A line perpendicular to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1: n$. Then, the equation of line is
Straight Lines
Solution:
Let the given points are $A(1,0)$ and $B(2,3)$. Let the line $P Q$ divide $A B$ in the ratio $1: n$ at $R$.
Using section formula for internal division
$R =\left(\frac{1 \times x_2+n \times x_1}{1+n}, \frac{1 \times y_2+n \times y_1}{1+n}\right) $
$ =\left(\frac{1 \times 2+n \times 1}{1+n}, \frac{1 \times 3+n \times 0}{1+n}\right) $
$ =\left(\frac{n+2}{n+1}, \frac{3}{1+n}\right) \left(\because x_1=1, y_1=0, x_2=2, y_2=3\right)$
Also, $P Q \perp A B$
Let slope of line $P Q$ is $m$.
$\therefore$ Slope of line $P Q \times$ Slope of line $A B=-1$
$(\because m_1m_2 = -1)$
$\Rightarrow m \times \frac{y_2-y_1}{x_2-x_1} =-1 $
$\Rightarrow m \times \frac{3-0}{2-1}=-1 $
$\Rightarrow m \times 3 =-1 $
$\Rightarrow m =-\frac{1}{3}$
Now, equation of line $P Q$ by using
$y-y_1=m\left(x-x_1\right)$
$ \Rightarrow y-\frac{3}{1+n}=\frac{-1}{3}\left(x-\frac{n+2}{n+1}\right) $
$ {\left[\because R\left(\frac{n+2}{n+1}, \frac{3}{1+n}\right)=\left(x_1, y_1\right)\right]}$
$\Rightarrow \frac{3(n+1) y-9}{1+n}=\frac{-x(n+1)+(n+2)}{n+1} $
$ \Rightarrow 3(n+1) y-9=-x(n+1)+(n+2) $
$ \Rightarrow x(n+1)+3(n+1) y=n+2+9$
$ \Rightarrow x(n+1)+3(n+1) y=n+11$
