Question
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.
✓
Answer
Let point C divides the join of A(1, 0) and B(2, 3) in the ratio 1 : n.
$\therefore$ Coordinates of C are $\left( {\frac{{2 + n}}{{1 + n}},\frac{3}{{1 + n}}} \right)$
Slope of AB $= \frac{{3 - 0}}{{2 - 1}} = 3$
Since the required line is perpendicular to AB,
$\therefore$ ;Slope of required line $m = - \frac{1}{3}$
Now the required line passing through point $\left( {\frac{{2 + n}}{{1 + n}},\frac{3}{{1 + n}}} \right)$ having slope $-\frac{1}{3}$.
$\therefore$ Equation of required line is
$y - \frac{3}{{1 + n}} = \frac{{ - 1}}{3}\left( {x - \frac{{2 + n}}{{1 + n}}} \right)$
$\Rightarrow \frac{{(1 + n)y - 3}}{{1 + n}} = - \frac{1}{3}\left[ {\frac{{(1 + n)x - 2 - n}}{{1 + n}}} \right]$
$\Rightarrow$ 3(1 + n)y - 9 = -(1 + n)x + 2 + n
$\Rightarrow$ (1+ n)x + 3 ( 1+ n) y = n + 11.
$\therefore$ Coordinates of C are $\left( {\frac{{2 + n}}{{1 + n}},\frac{3}{{1 + n}}} \right)$
Slope of AB $= \frac{{3 - 0}}{{2 - 1}} = 3$
Since the required line is perpendicular to AB,
$\therefore$ ;Slope of required line $m = - \frac{1}{3}$
Now the required line passing through point $\left( {\frac{{2 + n}}{{1 + n}},\frac{3}{{1 + n}}} \right)$ having slope $-\frac{1}{3}$.
$\therefore$ Equation of required line is
$y - \frac{3}{{1 + n}} = \frac{{ - 1}}{3}\left( {x - \frac{{2 + n}}{{1 + n}}} \right)$
$\Rightarrow \frac{{(1 + n)y - 3}}{{1 + n}} = - \frac{1}{3}\left[ {\frac{{(1 + n)x - 2 - n}}{{1 + n}}} \right]$
$\Rightarrow$ 3(1 + n)y - 9 = -(1 + n)x + 2 + n
$\Rightarrow$ (1+ n)x + 3 ( 1+ n) y = n + 11.
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