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

24. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

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alok kumar singh | Contributor-Level 10

4 months ago

24.

Slope of line passing through points (2, 5) and (–3, 6) is

$m_{1} = \frac{- 1}{m_{1}} = h$

$\frac{6 - 5}{- 3 - 2} = \frac{1}{- 5}$

So, slope of line perpendicular to line through (2, 5) and (–3, 6) is

$m_{2} = - \frac{1}{(\frac{- 1}{5})} = 5$

$∴$ Equation of line with slope 5 and passing through (–3, 5) is

$\Rightarrow 5 = \frac{y - 5}{x - (- 3)} = \frac{y - 5}{x + 3}$

$\Rightarrow 5 (x + 3) = y - 5$

$\Rightarrow 5 x + 15 - y + 5 = 0$

$\Rightarrow 5 x - y + 20 = 0$

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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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Eqⁿ : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct

If two circle x² + y² = 4 and x² + y² – 4lx + 9 = 0 intersect at two distinct points, then find the range of l.

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|r₁ – r₂| < c₁c₂ < r₁ + r₂

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

$| 2 λ | - 2 < \sqrt[]{4 λ^{2} - 9}$

$4 λ^{2} + 4 - 8 | λ | < 4 λ^{2} - 9$

$λ > \frac{13}{8}, λ < - \frac{13}{8}$

$\sqrt[]{4 λ^{2} - 9} > 0$

$λ > \frac{3}{2}, λ < - \frac{3}{2}$

$λ \in (- \infty, - \frac{13}{8}) \cup (\frac{13}{8}, \infty)$

Now,

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

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

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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

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

-> $m^{2} = \frac{1}{4}$

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

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

->Minimum = 18

The image of the point (3, 5) in the line x – y + 1 = 0, lies on:

Vishal Baghel

Kindly consider the following figure

A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of the line on the coordinate axes is $\frac{1}{4} .$ Three stones A, B and C are placed at the point (1, 1), (2, 2) and (4, 4) respectively. Then which of these stones is / are on the path of the man?

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According to question,

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

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

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

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24. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

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