COMEDK · Maths · 10. Straight Lines
Let the line \(L_1\) be a line passing through the point \((0, -6)\) and making an angle of \(150^\circ\) with the positive \(x\)–axis. Then the equation of a line \(L_2\) parallel to \(L_1\) and crossing the \(y\)–axis 2 units below the origin is:
- A \(x + \sqrt{3}y + 2\sqrt{3} = 0\)
- B \(x - \sqrt{3}y + 6\sqrt{3} = 0\)
- C \(x\sqrt{3} + y + 6 = 0\)
- D \(x - \sqrt{3}y - 2\sqrt{3} = 0\)
Answer & Solution
Correct Answer
(A) \(x + \sqrt{3}y + 2\sqrt{3} = 0\)
Step-by-step Solution
Detailed explanation
The slope of line \(L_1\) is \(m = \tan(150^\circ) = -\dfrac{1}{\sqrt{3}}\). Since line \(L_2\) is parallel to \(L_1\), the slope of \(L_2\) is also \(m = -\dfrac{1}{\sqrt{3}}\). Line \(L_2\) crosses the \(y\)-axis \(2\) units below the origin, so its \(y\)-intercept is…
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