Question

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

• A. $x(n+1)+3(n+1)y=n+11$
• B. $x(n+1)+(3n+1)=n+11$
• C. $x(n+1)-3(n+1)y=n+11$
• D. $x(n+1)-(3n+1)y=n+11$

Answer 1

Answer: (A)

Let the given points are $A(1,0)$ and $B(2,3)$. Let the line $PQ$ divide $AB$ 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, $PQ\perp AB$
Let slope of line $PQ$ is $m$.
$\therefore$ Slope of line $PQ\times$ Slope of line $AB=-1$
$(\because m_1{m}_2=-1)$
$\Rightarrow m\times \frac{y_2-y_1}{x_2-x_1}=-1$
$\Rightarrow m\times \frac{3-0}{2-1}=-1$
$\Rightarrow m\times 3=-1$
$\Rightarrow m=-\frac{1}{3}$
Now, equation of line $PQ$ by using
$y-y_1=m\left(x-x_1\right)$
$\Rightarrow y-\frac{3}{1+n}=\frac{-1}{3}\left(x-\frac{n+2}{n+1}\right)$
$\left[\because R\left(\frac{n+2}{n+1},\frac{3}{1+n}\right)=\left(x_1,y_1\right)\right]$
$\Rightarrow \frac{3(n+1)y-9}{1+n}=\frac{-x(n+1)+(n+2)}{n+1}$
$\Rightarrow 3(n+1)y-9=-x(n+1)+(n+2)$
$\Rightarrow x(n+1)+3(n+1)y=n+2+9$
$\Rightarrow x(n+1)+3(n+1)y=n+11$