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

21. Passing through the points (–1, 1) and (2, – 4).

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alok kumar singh | Contributor-Level 10

4 months ago

.Here (x₁, y₁) = (–1, 1) and (x₂, y₂) = (2, –4)

Using two point form, equation of line is

$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

$\Rightarrow y - 1 = \frac{- 4 - 1}{2 - (- 1)} (x - (- 1))$

$\Rightarrow y - 1 = \frac{- 5}{2 + 1} (x + 1)$

$\Rightarrow 3 (y - 1) = - 5 (x + 1)$

$\Rightarrow 3 y - 3 = - 5 x - 5$

$\Rightarrow 5 x + 3 y - 3 + 5 = 0$

$\Rightarrow 5 x + 3 y + 2 = 0$

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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

alok kumar singh

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.

alok kumar singh

|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,

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

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

$4 \sqrt[]{4 λ^{2} - 9} > - 5 \Rightarrow λ \in R$

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

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.

alok kumar singh

(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?

Vishal Baghel

According to question,

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

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

Equation of required line is $\frac{x}{a} + \frac{y}{b} = 1$ $\frac{}{} \frac{}{}$

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

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21. Passing through the points (–1, 1) and (2, – 4).

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