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)

$3y + x = \frac{n+11}{n+1}$

Answer 2

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
$\left(y - \frac{3}{ n+1}\right) = -\frac{1}{3}\left(x - \frac{2+n}{n+1}\right)$
$\Rightarrow 3y - \frac{9}{n+1} = -x + \frac{2+n}{n+1}$
$\Rightarrow 3y +x = \frac{11 + n}{n+1}$