COMEDK · Maths · 10. Straight Lines
The slope of lines which makes an angle \(60^{\circ}\) with the line \(y-3 x+18=0\)
- A \(\frac{3 \sqrt{3}-3}{1+3 \sqrt{3}}, \frac{3 \sqrt{3}-3}{1+3 \sqrt{3}}\)
- B \(\frac{3-\sqrt{3}}{1+3 \sqrt{3}}, \frac{3+\sqrt{3}}{1-3 \sqrt{3}}\)
- C \(\frac{3}{1+\sqrt{3}}, \frac{3}{1-\sqrt{3}}\)
- D \(\frac{\sqrt{3}-1}{3}, \frac{\sqrt{3}+1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{3-\sqrt{3}}{1+3 \sqrt{3}}, \frac{3+\sqrt{3}}{1-3 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Slope of the line, \(\begin{aligned} & y-3 x+18=0 \\ & \Rightarrow y=3 x-18 \Rightarrow \text { Slope }\left(m_1\right)=3 \end{aligned}\) and angle \((\theta)=60^{\circ}\)…
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