COMEDK · Maths · 10. Straight Lines
The slope of lines which makes an angle \(60^{\circ}\) with the line \(y-3 x+18=0\)
- A \(\dfrac{3}{1+\sqrt{3}}, \dfrac{3}{1-\sqrt{3}}\)
- B \(\dfrac{3 \sqrt{3}-3}{1+3 \sqrt{3}}, \dfrac{3 \sqrt{3}-3}{1+3 \sqrt{3}}\)
- C \(\dfrac{3-\sqrt{3}}{1+3 \sqrt{3}}, \dfrac{3+\sqrt{3}}{1-3 \sqrt{3}}\)
- D \(\dfrac{\sqrt{3}-1}{3}, \dfrac{\sqrt{3}+1}{3}\)
Answer & Solution
Correct Answer
(C) \(\dfrac{3-\sqrt{3}}{1+3 \sqrt{3}}, \dfrac{3+\sqrt{3}}{1-3 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
The given line is \(y - 3x + 18 = 0\), which can be written as \(y = 3x - 18\). The slope of this line is \(m_1 = 3\). Let the slope of the required lines be \(m\). The angle \(\theta\) between the two lines is given by…
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