Straight Lines

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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Contributor-Level 10

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

Option (B) is correct

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a week ago

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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Contributor-Level 10

|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,

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

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

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

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

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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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Contributor-Level 10

(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

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New answer posted

3 weeks ago

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

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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New answer posted

3 weeks ago

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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Vishal Baghel

Contributor-Level 10

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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3 weeks ago

If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x² + y² = 1 is a circle of radius r, then r is equal to:

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Contributor-Level 10

$h = \frac{c o s θ + 3}{2}$

$k = \frac{s i n θ + 2}{2}$

-> $c o s θ = 2 h - 3 & s i n θ = 2 k - 2$

-> ${(h - 3 / 2)}^{2} + {(k - 1)}^{2} = \frac{1}{4}$

$∴$ circle of radius r = $\frac{1}{2}$

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3 weeks ago

LE the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and y² = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e^{2}}$ is equal to…….

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Contributor-Level 9

Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (-3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

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Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (-3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

and length of latus rectum

$= \frac{2 b^{2}}{a} = 3$

Hence $\frac{l}{e^{2}} = \frac{3}{3 / 4} = 4$

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New answer posted

3 weeks ago

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2 a n d l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2} a n d l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4} i s :$

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Contributor-Level 9

$? l_{1} a n d l_{2}$ are perpendicular, so

$3 * 1 + (- 2) (\frac{α}{2}) + 0 * 2 = 0$

⇒a = 3

Now angle between $l_{2} a n d l_{3}$ ,

$c o s θ = \frac{1 (- 3) + \frac{α}{2} (- 2) + 2 (4)}{\sqrt[]{1 + \frac{α^{2}}{4} + 4} . \sqrt[]{9 + 4 + 16}}$

$\Rightarrow c o s θ = \frac{\frac{2}{29}}{2} \Rightarrow θ = c o s^{- 1} (\frac{4}{29}) = s e c^{- 1} (\frac{29}{4})$

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New answer posted

3 weeks ago

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $Δ P Q R$ is:

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Contributor-Level 9

RM = $| \frac{3 + 7 - 5}{\sqrt[]{2}} | = \frac{5}{\sqrt[]{2}}$

$l s i n 60 ° = \frac{5}{\sqrt[]{2}} \Rightarrow l = 5 \sqrt[]{\frac{2}{3}}$

$∴ A r e a o f Δ P Q R = \frac{\sqrt[]{3}}{4} l^{2} = \frac{25}{2 \sqrt[]{3}}$

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New answer posted

3 weeks ago

Let the lines y + 2x = $\sqrt[]{11} + 7 \sqrt[]{7}$ and $2 y + x = 2 \sqrt[]{11} + 6 \sqrt[]{7}$ be normal to a circle C : ${(x - h)}^{2} + {(y - k)}^{2} = r^{2} .$ If the line $\sqrt[]{11} y - 3 x = \frac{5 \sqrt[]{77}}{3} + 11$ is tangent to the circle C, then the value of $\sqrt[]{11} y - 3 x = \frac{5 \sqrt[]{77}}{3} + 11$ is equal to…………….

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