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The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line 3 x+y = 1 is ______. - Mathematics

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Question

The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line `sqrt(3)  x + y` = 1 is ______.

Options

  • y + 2 = 0, `sqrt(3) x - y - 2 - 3 sqrt(3)` = 0

  • x – 2 = 0, `sqrt(3)x - y + 2 + 3 sqrt(3)` = 0

  • `sqrt(3) x - y - 2 - 3sqrt(3)` = 0

  • None of these

MCQ
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Solution

The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line `sqrt(3)  x + y` = 1 is y + 2 = 0, `sqrt(3) x - y - 2 - 3 sqrt(3)` = 0.

Explanation:

Equation of line is given by `sqrt(3)x + y + 1` = 0

`sqrt(3)x + y = 1` = 0

⇒ y = `- sqrt(3)x - 1`

∴ Slope of this line, m1  `- sqrt(3)`

Let m2 be the slope of the required line

∴ tan θ = `|(m_1 - m_2)/(1 + m_1m_2)|`

⇒ tan 60° = `|(-sqrt(3) - m_2)/(1 + (- sqrt(3))m_2)|`

⇒ `sqrt(3) = +-  ((- sqrt(3) - m_2)/(1 - sqrt(3)m_2))`

⇒ `sqrt(3) = (- sqrt(3) - m_2)/(1 - sqrt(3)m_2)`  ....[Taking (+) sign]

⇒ `sqrt(3) - 3m_2 = - sqrt(3) -m_2`

⇒ `2m_2 = 2sqrt(3)`

⇒ m2 = `sqrt(3)`

And `sqrt(3) - ((- sqrt(3) - m_2)/(1 - sqrt(3)m_2))` ....[Taking (–) sign]

⇒ `sqrt(3) = (sqrt(3) + m_2)/(1 - sqrt(3)m_2)`

⇒ `sqrt(3) - 3m_2 = sqrt(3) + m_2`

⇒ 4m2 = 0

⇒ m2 = 0

∴ Equation of line passing through (3, – 2) with slope `sqrt(3)` is

y + 2 = `sqrt(3)(x - 3)`

⇒ y + 2 = `sqrt(3)x - 3sqrt(3)`

⇒ `sqrt(3)x - y - 2 - 3sqrt(3)` = 0

And the equation of line passing through (3, –2) with slope 0 is

y + 2 = 0(x – 3)

⇒ y + 2 = 0

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Equation of a Straight Line - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
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Chapter 10: Straight Lines - Exercise [Page 181]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 10 Straight Lines
Exercise | Q 29 | Page 181

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