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

49. Find the equation of the right bisector of the line segment joining the points (3, 4)and (–1, 2).

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alok kumar singh | Contributor-Level 10

4 months ago

49.

Let P(3, 9) and Q (-1, 2) be the point. Let M bisects PQ at M so,

Co-ordinate of M = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

$= (\frac{3 + (- 1)}{2}, \frac{4 + 2}{2})$

$= (\frac{3 - 1}{2}, \frac{6}{2}) = (\frac{2}{2}, \frac{6}{2}) = (1, 3)$

Now, slope of AB, m = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - 4}{- 1 - 3} = \frac{- 2}{- 4} = \frac{1}{2} .$

As the bisects AB perpendicular it has slope $- \frac{1}{m}$ and it passes through M(1, 3) it has the equation of the form,

$\frac{- 1}{m} = \frac{y - y_{σ}}{x - x_{0}}$

$\frac{- 1}{\frac{1}{2}} = \frac{y - 3}{x - 1}$

$\Rightarrow - 2 = \frac{y - 3}{x - 1}$

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

$\Rightarrow - 2 x + 2 = y - 3$

$\Rightarrow 2 x + y - 3 - 2 = 0$

2x + y - 5 = 0

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A line passes through (9, 0), making angle 30° with positive direction of x-axis. It is rotated by angle of 15° with respect to (9, 0). Then one of the equation of new line is

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|r₁ – r₂| < c₁c₂ < r₁ + r₂

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

$| 2 λ | - 2 < \sqrt[]{4 λ^{2} - 9}$

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The line passes through the centre of circle x² + y² – 16x – 4y = 0, it interacts with the positive coordinate axis at A & B. Then find the minimum value of OA + OB, where O is origin.

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(y – 2) = m (x – 8)

⇒ x-intercept

⇒ $(\frac{- 2}{m} + 8)$

⇒ y-intercept

⇒ (–8m + 2)

⇒ OA + OB = $\frac{- 2}{m^{2}}$ + 8 – 8m + 2

$f' (m) = \frac{2}{m^{2}} - 8 = 0$

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-> $m = \frac{- 1}{2}$

-> $f (\frac{- 1}{2}) = 18$

->Minimum = 18

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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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49. Find the equation of the right bisector of the line segment joining the points (3, 4)and (–1, 2).

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