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
$P\equiv\left(\frac{2\times1+1\times n}{n+1}, \frac{3\times1+0\times n}{n+1}\right)$
$[$ 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} $