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

A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = $\sqrt[]{5}$ inches and $∠ P C B = t a n^{- 1} (2) .$

The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is:

Option 1 -

$t a n^{- 1} (\frac{1}{2})$

Option 2 -

$t a n^{- 1} (\frac{4}{3})$

Option 3 -

$t a n^{- 1} (1)$

Option 4 -

$t a n^{- 1} (\frac{3}{4})$

5 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$t a n^{- 1} 2 = θ$

->tan q = 2

$s i n (θ - α) = \frac{1}{\sqrt[]{5}}$

->4 – 2 tanq = 1 + 2 tan a tan a = $\frac{3}{4}$

$α = t a n^{- 1} \frac{3}{4}$

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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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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

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

$| 2 λ | - 2 < \sqrt[]{4 λ^{2} - 9}$

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$λ > \frac{3}{2}, λ < - \frac{3}{2}$

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

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

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

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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,

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

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

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