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

Let ABC be a triangle with A (-3, 1) and $∠ A C B = θ,$ $0 < θ < \frac{π}{2} .$ If the equation of the median through B is 2x + y – 3 = 0 and the equation of angle bisector of C is 7x – 4y – 1 = 0, then tanq is equal to:

Option 1 -

$\frac{4}{3}$

Option 2 -

$\frac{3}{4}$

Option 3 -

$\frac{1}{2}$

Option 4 -

3 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

Line BM : 2x + y = 3 Þ M (0, 3)

Line CD : 7x – 4y = 1 Þ C (3, 5)

Mirror image of A (-3, 1) in the line CD is $(\frac{13}{5}, \frac{- 11}{5})$ and it will lie on BC.

Slope of AC is 2/3

Slope of BC is 18.

$t a n θ = \frac{18 - \frac{2}{3}}{1 + 18 (\frac{2}{3})} = \frac{4}{3}$

where θ = $∠ A C B$

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-> $| 2 - \sqrt[]{4 λ^{2} - 9} | < | 2 λ | < 2 + \sqrt[]{4 λ^{2} - 9}$

$| 2 λ | - 2 < \sqrt[]{4 λ^{2} - 9}$

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⇒ $(\frac{- 2}{m} + 8)$

⇒ y-intercept

⇒ (–8m + 2)

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Kindly consider the following figure

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According to question,

$\frac{1}{2} (\frac{1}{a} + \frac{1}{b}) = \frac{1}{4}$

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Obviously B (2, 2) satisfying condition (i)

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Let ABC be a triangle with A (-3, 1) and $∠ A C B = θ,$ $0 < θ < \frac{π}{2} .$ If the equation of the median through B is 2x + y – 3 = 0 and the equation of angle bisector of C is 7x – 4y – 1 = 0, then tanq is equal to:

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Let ABC be a triangle with A (-3, 1) and ∠ A C B = θ , 0 < θ < π 2 . If the equation of the median through B is 2x + y – 3 = 0 and the equation of angle bisector of C is 7x – 4y – 1 = 0, then tanq is equal to:

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Let ABC be a triangle with A (-3, 1) and $∠ A C B = θ,$ $0 < θ < \frac{π}{2} .$ If the equation of the median through B is 2x + y – 3 = 0 and the equation of angle bisector of C is 7x – 4y – 1 = 0, then tanq is equal to: