<p><strong>61. Find the equation of a line drawn perpendicular to the line <math><mrow><mfrac><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>y</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>=</mo><mn>1</mn></mrow></math> through the point, where it meets the y-axis.</strong></p>

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

61. Find the equation of a line drawn perpendicular to the line $\frac{x}{3} + \frac{y}{6} = 1$ through the point, where it meets the y-axis.

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alok kumar singh | Contributor-Level 10

4 months ago

61. The given Eqⁿ of the line is $\frac{x}{4} + \frac{y}{6}$ = 1 ______ (1)

so, Slope of line = - $\frac{6}{4} = - \frac{3}{2} .$

The line ⊥ to line (1) say l₂ has

Slope of l₂ = $\frac{- 1}{(- 3 / 2)} = \frac{2}{3} .$

Let P (0, y) be the point of on y-axis where it is cut by the line (1)

Then, $\frac{0}{4} + \frac{y}{6} = 1$

y = 6

i.e, the point P has co-ordinate (0, 6)

Eqⁿ of line ⊥ to $\frac{x}{4} + \frac{y}{6} = 1$ and cuts y-axis at P (0,6) is

y – 6 = $\frac{2}{3}$ (x – 0)

3y – 18 = 2x

2x – 3y + 18 = 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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