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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
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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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