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

29. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

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alok kumar singh | Contributor-Level 10

4 months ago

29.

The slope of line passing through (0, 0) and (–2, 9) is

$m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{9 - 0}{- 2 - 0} = \frac{9}{- 2}$

The line perpendicular to the line having slope m1 will have the slope

$m_{2} = \frac{- 1}{m_{1}} = \frac{- 1}{- 9 / 2} = \frac{2}{9}$

So, the equation of line with slope m2 and passing $(x_{0}, y_{0}) = (- 2, 9)$ is

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

$\Rightarrow \frac{2}{9} = \frac{y - 9}{x - (- 2)} = \frac{y - 9}{x + 2}$

$\Rightarrow 2 (x + 2) = 9 (y - 9)$

$\Rightarrow 2 x + 4 = 9 y - 81$

$\Rightarrow 2 x - 9 y + 4 + 81 = 0$

$\Rightarrow 2 x - 9 y + 85 = 0$

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29. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

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