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Ncert Solutions Maths class 11th Maths Class 11th Straight Lines

25. 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.

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alok kumar singh | Contributor-Level 10

4 months ago

25.

Let P(1, 0) and Q(2, 3) be the given points. Then, slope of line joining PQ,

$m_{1} = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3$

If A divides PQ with ratio 1:n, co-ordinate of A is

$(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$

$= (\frac{1 \times 2 + n \times 1}{1 + n}, \frac{1 \times 3 + n \times 0}{1 + n})$

$= (\frac{2 + n}{1 + n}, \frac{3}{1 + n})$

So, slope of line perpendicular to line joining points P and Q

$m_{2} = \frac{- 1}{m_{1}} = - \frac{1}{3}$

Hence, equation of line passing through point A and (perpendicular to line) joining points P and Q is given by

$m_{2} = \frac{y - y_{0}}{x - x_{0}}$

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

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

$\Rightarrow - [x (n + 1) - (2 + n)] = 3 [y (n + 1) - 3]$

$\Rightarrow - x (n + 1) + (2 + n) = 3 y (n + 1) - 3 \times 3$

$\Rightarrow x (n + 1) + 3 (n + 1) y - (2 + n) - 9 = 0$

$\Rightarrow x (n + 1) + 3 (n + 1) y - n - 2 - 9 = 0$

$\Rightarrow x (n + 1) + 3 (n + 1) y - n - 11 = 0$

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25. 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.

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