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TS EAMCET · Maths · Straight Lines

The equation of the straight line passing through the point \((3,2)\) and inclined at angle of \(60^{\circ}\) with the line \(\sqrt{3} x+y=1\) is

  1. A \(\sqrt{3} x+y-(2+3 \sqrt{3})=0\)
  2. B \(\sqrt{3} x-y+(2-3 \sqrt{3})=0\)
  3. C \(-\sqrt{3} x+y-(2-3 \sqrt{3})=0\)
  4. D \(-\sqrt{3} x+y+(2-3 \sqrt{3})=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{3} x-y+(2-3 \sqrt{3})=0\)

Step-by-step Solution

Detailed explanation

\[ \begin{aligned} & \sqrt{3} x+y=1 \\ & m_1=-\sqrt{3} \end{aligned} \] Let slope of required line be \(m_2\) \[ \begin{aligned} & \tan 60^{\circ}=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ & \Rightarrow \sqrt{3}=\left|\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\right| \end{aligned} \]…
From TS EAMCET
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