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
- A \(\sqrt{3} x+y-(2+3 \sqrt{3})=0\)
- B \(\sqrt{3} x-y+(2-3 \sqrt{3})=0\)
- C \(-\sqrt{3} x+y-(2-3 \sqrt{3})=0\)
- D \(-\sqrt{3} x+y+(2-3 \sqrt{3})=0\)
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 \(m_2\) \[ \begin{aligned} & \tan 60^{\circ}=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ & \Rightarrow \sqrt{3}=\left|\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\right| \end{aligned} \]…
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