Question

A line perpendicular to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1:n$ . The equation of the line is

• A. $3y+x=\frac{n+11}{n+1}$
• B. $3y-x=\frac{n+11}{n+1}$
• C. $3y+x=\frac{n-11}{n+1}$
• D. $3y-x=\frac{n+11}{n-1}$

Answer 1

Answer: (A)

Slope of $AB=\frac{3-0}{2-1}=\frac{3}{1}$

$\therefore$ Slope of $MN=-\frac{1}{3}$
Now, coordinates of
$P\equiv \left(\frac{2\times 1+1\times n}{n+1},\frac{3\times 1+0\times n}{n+1}\right)$
$[$ By section formulaa$]$
$\equiv \left(\frac{2+n}{n+1},\frac{3}{n+1}\right)$
$\therefore$ Required equation is given as
$\left(y-\frac{3}{n+1}\right)=-\frac{1}{3}\left(x-\frac{2+n}{n+1}\right)$
$\Rightarrow 3y-\frac{9}{n+1}=-x+\frac{2+n}{n+1}$
$\Rightarrow 3y+x=\frac{11+n}{n+1}$