Let \( \alpha \)1, \( \alpha \)2 (\( \alpha \)1 < \( \alpha \)2) be the values of \( \alpha \) fo the points (\( \alpha \), \( - \)3), (2, 0) and (1, \( \alpha \)) to be collinear. Then the equation of the line, passing through (\( \alpha \)1, \( \alpha \)2) and making an angle of \( {\pi \over 3} \) with the positive direction of the x-axis, is :
Let \( \alpha \)1, \( \alpha \)2 (\( \alpha \)1 < \( \alpha \)2) be the values of \( \alpha \) fo the points (\( \alpha \), \( - \)3), (2, 0) and (1, \( \alpha \)) to be collinear. Then the equation of the line, passing through (\( \alpha \)1, \( \alpha \)2) and making an angle of \( {\pi \over 3} \) with the positive direction of the x-axis, is :
Show Hint
- \( x - \sqrt 3 y - 3\sqrt 3 + 1 = 0 \)
- \( \sqrt 3 x - y + \sqrt 3 + 3 = 0 \)
- \( x - \sqrt 3 y + 3\sqrt 3 + 1 = 0 \)
- \( \sqrt 3 x - y + \sqrt 3 - 3 = 0 \)
The Correct Option is B
Solution and Explanation
Step 2: \(\alpha=3\) or \(-1\); with \(\alpha_1<\alpha_2\): \(\alpha_1=-1,\ \alpha_2=3\).
Step 3: Line through \((-1,3)\) with slope \(\tan\tfrac{\pi}{3}=\sqrt3\): \(y-3=\sqrt3(x+1)\Rightarrow \sqrt3x-y+\sqrt3+3=0\).
Answer: (B) \(\sqrt3x-y+\sqrt3+3=0\)
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