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

27. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

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alok kumar singh | Contributor-Level 10

4 months ago

Let a and b be the intercepts on x and y-axis

Then a + b = 9

$\Rightarrow b = 9 - a (1)$

Using intercept form equation of line with a and b intercepts are

$\frac{x}{a} + \frac{y}{b} = 1$

$\Rightarrow \frac{x}{a} + \frac{y}{9 - a} = 1 (2)$

As point (2, 2) passes the line having equation of form equation (2) we can write as

$\frac{2}{a} + \frac{2}{9 - a} = 1$

$\Rightarrow \frac{2 (9 - a) + 2 a}{a (9 - a)} = 1$

$\Rightarrow 18 - 2 a + 2 a = (9 - a) a$

$\Rightarrow 18 = 9 a - a^{2}$

$\Rightarrow a^{2} - 9 a + 18 = 0$

$\Rightarrow a^{2} - 3 a - 6 a + 18 = 0$

$\Rightarrow a (a - 3) - 6 (a - 3) = 0$

$\Rightarrow (a - 3) (a - 6) = 0$

So, a = 3, 6

Case I

When a = 3, b = 9 – 3 = 6. Then the equation of line is

$\frac{x}{3} + \frac{y}{c} = 1$

$\Rightarrow \frac{2 x + y}{6} = 1$

$\Rightarrow 2 x + y = 6$

$\Rightarrow 2 x + y - 6 = 0$

Case II

When a = 6, b = 9 – 6 = 3. Then equation of line is

$\frac{x}{6} + \frac{y}{3} = 1$

$\Rightarrow \frac{x + 2 y}{6} = 1$

$\Rightarrow x + 2 y = 6$

$\Rightarrow x + 2 y - 6 = 0$

Hence, equation of line is 2x + y – 6 or x + 2y – 6 = 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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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,

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

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

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

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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:

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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?

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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27. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

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